Waves Past Papers

Concept:

Amplitude is a fundamental spatial parameter of a periodic wave, defined as the maximum displacement of any point on the wave from its equilibrium (mean) position.

Formula / Rule / Reaction:

$$\text{Amplitude } (A) = \frac{\text{Peak-to-Trough Vertical Height}}{2}$$

Solution:

• By definition, amplitude is measured from the central resting equilibrium line (\(y = 0\)) up to the peak of a crest or down to the bottom of a trough.



• In the given wave diagram, the total vertical distance from the bottom of the trough to the peak of the crest is \( 2\text{ m} \).



• Because the equilibrium line divides this peak-to-trough height equally in two:



• $$A = \frac{2\text{ m}}{2} = 1\text{ m}$$

Why other options are incorrect:

• Option A (2 m): This is the total peak-to-trough vertical span (total height), which is twice the true amplitude. Confusing peak-to-trough height with amplitude is a classic exam trap.
• Option B (1/2 m) & Option C (\( \frac{\sqrt{3}}{2} \text{ m} \)): Incorrect numerical values that do not correspond to half the measured vertical distance.

Concept:

In Simple Harmonic Motion, maximum acceleration occurs at the points of maximum displacement (the amplitude), where the restoring force is strongest.

Formula:

$$ F = ma $$
$$ F = kx $$
$$ a = \frac{kx}{m} $$

Solution:

• We are given: Mass \( m = 10 \text{ kg} \), Spring constant \( k = 20 \text{ N/m} \).



• The maximum displacement (amplitude) is given in centimeters. Convert it to standard SI units (meters): \( x = 20 \text{ cm} = 0.2 \text{ m} \).



• Plug the values into the acceleration formula:



• \( a = \frac{20 \times 0.2}{10} \).



• \( a = \frac{4}{10} = 0.4 \text{ m s}^{-1} \).

Why other options are incorrect:

Option B (4) happens if a student forgets to divide by the mass (10 kg). Option A and C result from not converting centimeters to meters.

Concept:

Wavelength is inversely proportional to frequency. Therefore, the lowest frequency sound will mathematically yield the longest (maximum) wavelength.

Formula:

$$ \lambda = \frac{v}{f} $$

Solution:

• To find the maximum wavelength, we must use the minimum audible frequency, which is \( f = 20 \text{ Hz} \).



• Assume the standard speed of sound in room-temperature air is roughly \( v \approx 340 \text{ m/s} \).



• Substitute the values: \( \lambda_{max} = \frac{340}{20} \).



• \( \lambda_{max} = 17 \text{ m} \).

Why other options are incorrect:

Options B (17 mm) and C (17 cm) correspond to the high-frequency limit (20,000 Hz), which yields the minimum, not maximum, wavelength. 17 km is completely out of scale.

Concept:

A wave's structural geometry divides neatly into fractions of a total wavelength.

Solution:

• One complete wavelength (\( \lambda \)) is the full distance from one crest to the very next crest.



• Because standard waves are symmetrical, a trough sits exactly halfway between two consecutive crests.



• Therefore, the horizontal distance from the peak of a crest to the deepest point of the immediately adjacent trough is exactly half of a full wavelength: \( \lambda/2 \).

Why other options are incorrect:

\( \lambda \) is the distance from crest to crest. \( \lambda/4 \) is the horizontal distance from a crest to the zero-crossing equilibrium line.

Concept:

Frequency is defined as the total number of repetitive cycles (or pulses) generated per unit of time.

Formula:

$$ f = \frac{N}{t} $$

Solution:

• Given: Number of pulses \( N = 500 \) and total time \( t = 50 \text{ s} \).



• Substitute the values: \( f = \frac{500}{50} \).



• \( f = 10 \text{ Hz} \).



• Note that Hertz (Hz) is dimensionally equivalent to inverse seconds (\( \text{s}^{-1} \)). Thus, the answer is 10 \( \text{s}^{-1} \).

Why other options are incorrect:

Option B (0.1) is the Time Period (50/500), incorrectly inverted. Option C and D use raw numbers from the prompt without solving.

Concept:

The velocity of sound in a gas increases linearly with temperature over small ranges.

Formula:

$$ v_t = v_0 + 0.61t $$

Solution:

• We know the speed at \( 0^\circ\text{C} \) is \( v_0 = 332 \text{ m/s} \).



• The temperature rise is \( t = 10^\circ\text{C} \).



• Calculate the increment: \( 0.61 \times 10 = 6.1 \text{ m/s} \).



• Add this to the baseline speed: \( v_{10} = 332 + 6.1 = 338.1 \text{ ms}^{-1} \).

Why other options are incorrect:

Option C assumes the increment is 0.61 total, ignoring the multiplier of 10 degrees. Option A assumes no change. Option D is mathematically arbitrary.

Concept:

Frequency measures how often an event repeats. Terminology slightly differs based on whether we measure strictly cycles (Hz) or rotational phase (radians).

Solution:

• Note: Strictly speaking, "revolutions per second" defines standard rotational/linear frequency (\( f \), measured in Hz). Angular frequency (\( \omega \)) is technically radians per second (\( \omega = 2\pi f \)).



• However, in many historical exam contexts, rotational "revolutions" are loosely grouped under the broader umbrella of Angular frequency to distinguish it from linear back-and-forth oscillations, or it is treated as a synonym for rotational speed (\( \text{rpm/rps} \)). Based on the official exam key, Option B is considered the targeted answer.

Why other options are incorrect:

Time period is seconds per revolution. Amplitude is a spatial distance. Linear frequency is heavily debated here, but the official historical key selects B.

Concept:

Sound velocity depends on the elastic and inertial properties of the gas.

Formula:

$$ v = \sqrt{\frac{\gamma P}{\rho}} $$

Solution:

• According to Boyle's law, at a constant temperature, any increase in gas pressure (\( P \)) results in an exactly proportional increase in its density (\( \rho \)).



• Because both variables increase by the same factor, the ratio \( P/\rho \) remains perfectly constant.



• Therefore, changing the pressure has absolutely no effect on the speed of sound.

Why other options are incorrect:

Temperature dramatically alters speed. Moisture (humidity) lowers the average molecular mass of air, decreasing density and thus increasing sound speed.

Concept:

In Simple Harmonic Motion (like a simple pendulum or an ideal spring), the oscillation timing is defined by system constants.

Formula:

$$ T = 2\pi\sqrt{\frac{L}{g}} \quad \text{or} \quad T = 2\pi\sqrt{\frac{m}{k}} $$

Solution:

• Notice that neither formula contains the variable for amplitude (maximum displacement).



• This remarkable property is called isochronism: whether the body swings slightly or a bit more widely, it takes the exact same amount of time to complete one full cycle.



• Therefore, the time period is independent of Amplitude.

Why other options are incorrect:

Time period is the exact mathematical inverse of frequency (\( T = 1/f \)), making it entirely dependent on it. It also clearly depends on 'g' for pendulums.

Concept:

The frequency of a stretched string is determined by the wave speed, which in turn depends on the string's tension.

Formula:

$$ f \propto \sqrt{T} $$

Solution:

• The original frequency is \( f_1 \propto \sqrt{T} \).



• We want the new frequency \( f_2 \) to be exactly double: \( f_2 = 2f_1 \).



• To mathematically bring a factor of 2 outside the square root, the term inside must be multiplied by \( 2^2 = 4 \).



• Therefore, \( \sqrt{4T} = 2\sqrt{T} \).



• The new tension must be 4T.

Why other options are incorrect:

If tension is 2T, the frequency increases by \( \sqrt{2} \approx 1.41 \), not doubled. Tension of 16T would quadruple the frequency.

Concept:

The speed of a longitudinal mechanical wave (like sound) depends on the elasticity and density of the medium.

Solution:

• In solids, atomic bonds are incredibly rigid (high bulk/shear modulus), allowing energy to transfer very rapidly.



• In gases, molecules are spread far apart and must physically travel through empty space to collide with neighbors, making energy transfer slow.



• Therefore, longitudinal waves travel fastest in solids, intermediate in liquids, and the slowest in gases.



• They travel more slowly in Gases than in Solids.

Why other options are incorrect:

Options A and B contradict the established hierarchy of wave speeds across states of matter.

Concept:

The "time required for each new whole cycle" is the definition of the wave's Time Period (\( T \)).

Formula:

$$ v = \frac{\lambda}{T} \implies T = \frac{\lambda}{v} $$

Solution:

• The distance between adjacent crests gives us the wavelength: \( \lambda = 5.0 \text{ cm} \).



• The velocity is given as \( v = 25 \text{ cm/s} \).



• Since both units are in centimeters, we don't need to convert to meters.



• Substitute values: \( T = \frac{5.0}{25} \).



• \( T = \frac{1}{5} = 0.2 \text{ sec} \).

Why other options are incorrect:

Option B (4) is the frequency (\( f = v/\lambda = 25/5 = 5 \text{ Hz} \) ... wait, \( 25/5 = 5 \), so frequency is 5 Hz. \( 1/5 = 0.2 \)). Option A (100) comes from wrongly multiplying the terms instead of dividing.

Concept:

Standing waves alternate between points of zero movement (nodes) and points of maximum movement (antinodes).

Solution:

• A full wavelength (\( \lambda \)) contains two complete structural loops.



• The distance spanning a single complete loop (from node to consecutive node) is \( \lambda/2 \).



• An antinode is located at the exact physical center of a loop.



• Therefore, the distance from the edge of the loop (a node) to its center (an antinode) is half of \( \lambda/2 \), which equals \( \lambda/4 \).

Why other options are incorrect:

\( \lambda/2 \) represents the distance between two nodes or two antinodes. \( \lambda \) represents the length of two complete loops.

Concept:

When a string clamped at both ends vibrates, it forms a standing wave composed of discrete segments (loops). Each individual segment corresponds to exactly half a wavelength.

Formula:

$$ L = n \left( \frac{\lambda}{2} \right) $$

Solution:

• We are given the total length \( L = 150 \text{ cm} \) and the number of loops \( n = 4 \).



• Set up the equation: \( 150 = 4 \left( \frac{\lambda}{2} \right) \).



• Simplify the right side: \( 150 = 2\lambda \).



• Solve for wavelength: \( \lambda = \frac{150}{2} = 75 \text{ cm} \).

Why other options are incorrect:

Option A (33.3 cm) occurs if one divides 150 by 4, incorrectly assuming one loop = one full wavelength. Option C assumes wavelength equals the string length.

Concept:

The generic formula for the velocity of any mechanical wave through a medium depends on its elastic and inertial properties.

Formula:

$$ v = \sqrt{\frac{E}{\rho}} \quad \text{and} \quad v = f \lambda $$

Solution:

• The expression "square root of the ratio of elasticity (\( E \)) to mass density (\( \rho \))" is precisely the formula for Wave Velocity (\( v \)).



• According to the universal wave equation, wave velocity is also equal to the product of its frequency (\( f \)) and wavelength (\( \lambda \)).



• Therefore, \( \sqrt{E/\rho} = f\lambda \).

Why other options are incorrect:

Options A, C, and D are entirely different physical quantities with vastly different dimensional units. None of them evaluate to \( m/s \).

Concept:

The velocity of sound in an ideal gas relies on the ratio of pressure to density.

Formula:

$$ v = \sqrt{\frac{\gamma P}{\rho}} $$

Solution:

• When gas is enclosed in a tube and pressure (\( P \)) is increased 10 times at a constant temperature, Boyle's Law dictates that the volume shrinks, making the gas strictly 10 times denser.



• The new velocity would be \( v' = \sqrt{\frac{\gamma (10P)}{(10\rho)}} \).



• The 10s perfectly cancel out, leaving the original ratio intact.



• Thus, the speed of sound remains constant regardless of ambient pressure changes.

Why other options are incorrect:

Students often glance at the formula, see \( P \) in the numerator, and wrongly assume an increase (Option A) without accounting for the hidden proportional variable \( \rho \).

Concept:

The Doppler effect is utilized to detect relative velocity by measuring wave frequency shifts.

Solution:

• Note: While police radar inherently measures automobile speed (Option B), "Sending radar signals" (Doppler Radar) is technically the broader, foundational application listed in historical syllabi.



• Doppler Radar works by sending radar signals (microwaves) at a target. When the waves bounce off a moving target (like a storm front or a car), their frequency shifts.



• The system calculates the target's speed strictly based on this returning Doppler shift. According to the official exam key, Option D is the accepted answer, capturing the foundational mechanism.

Why other options are incorrect:

Holograms rely on wave interference. Determining the speed of light relies on classical timing experiments or interferometry, not the Doppler effect.

Concept:

All wave types share universal wave behaviors, while some behaviors are geometry-specific.

Solution:

• The principle of superposition is a universal wave property. It dictates that when two or more waves (whether transverse or longitudinal) overlap in space, their net displacement is the sum of their individual displacements.



• Sound waves (longitudinal) frequently undergo superposition to create interference patterns, beats, and standing waves.

Why other options are incorrect:

Option A (Polarization) is strictly exclusive to transverse waves, making it an excellent distractor. Longitudinal waves cannot be polarized because their oscillation is already locked to a single 1D axis.

Concept:

The universal wave equation relates the spatial, temporal, and kinematic properties of a wave.

Formula:

$$ v = f \lambda $$

Solution:

• Frequency (\( f \)) represents the number of wave cycles passing a point per second.



• Wavelength (\( \lambda \)) represents the physical length of one complete cycle.



• By multiplying them, you calculate the total distance the wave energy travels in one second, which is the definition of the Speed of the wave (\( v \)).

Why other options are incorrect:

Displacement and amplitude are vertical measurements of the medium. Time period is the inverse of frequency (\( 1/f \)), completely unrelated to this product.

Concept:

The geometry of a standing wave consists of repeating discrete segments or "loops".

Solution:

• A full wavelength (\( \lambda \)) must contain exactly two identical loops to complete a full cycle (one "up" loop and one "down" loop).



• The boundaries of a single loop are nodes. The center of a single loop is an antinode.



• Therefore, the physical distance spanning exactly one loop—whether measured from node to consecutive node, or antinode to consecutive antinode—is half of a wavelength: \( \lambda/2 \).

Why other options are incorrect:

\( \lambda/4 \) is the shortest distance between a node and an adjacent antinode. \( \lambda \) is the distance of two full loops.

Concept:

A string fixed at both ends produces harmonic frequencies that are pure integer multiples of its fundamental (lowest) frequency.

Formula:

$$ f_n = n f_1 $$

Solution:

• We are given the fundamental frequency (first harmonic), \( f_1 = 50 \text{ Hz} \).



• The question asks for the fourth harmonic, meaning \( n = 4 \).



• Substitute the values: \( f_4 = 4 \times 50 \text{ Hz} \).



• \( f_4 = 200 \text{ Hz} \).

Why other options are incorrect:

Option B (150) is the 3rd harmonic. Option D (250) is the 5th harmonic.

Concept:

Oscillatory terminology specifically differentiates between instantaneous position and peak limits.

Solution:

• At any given moment, the body's distance from the center is its instantaneous displacement.



• However, the absolute furthest point it reaches before turning back—the "maximum displacement" from equilibrium—is formally defined as the Amplitude.



• Amplitude determines the total energy housed within the vibrating system.

Why other options are incorrect:

Distance and displacement are variable measurements that change continuously during the swing. Frequency relates to time, not spatial boundaries.

Concept:

The velocity of sound in gases climbs predictably as ambient temperature rises.

Formula:

$$ v_t = v_0 + 0.61t $$

Solution:

• We are given the baseline speed \( v_0 = 332 \text{ m/s} \) and temperature \( t = 50^\circ\text{C} \).



• First, calculate the temperature-induced speed increment: \( 0.61 \times 50 \).



• \( 0.61 \times 50 = 30.5 \text{ m/s} \).



• Add this increment to the baseline: \( v_{50} = 332 + 30.5 = 362.5 \text{ m/s} \).

Why other options are incorrect:

Option D occurs if a math error yields 50 instead of 30.5. Option B assumes temperature has no effect on wave speed.

Concept:

Wavelength is strictly defined as the physical distance required for one complete wave cycle (one full crest and one full trough).

Solution:

• The provided diagram features an arrow spanning a total distance of exactly 2 meters.



• By tracing the wave from start to finish within that 2-meter span, we count exactly two identical, complete wave cycles (up-down, up-down).



• To find the length of a single cycle (wavelength, \( \lambda \)), divide the total distance by the number of cycles: \( \lambda = \frac{2 \text{ m}}{2 \text{ cycles}} \).



• Therefore, \( \lambda = 1 \text{ m} \).

Why other options are incorrect:

Option A (2m) is the trap for students who assume the provided spatial label automatically represents exactly one wavelength without counting the underlying cycles.

Concept:

The velocity of sound in gases relies heavily on the gas's internal restoring forces and mass, expressed mathematically by Laplace.

Formula:

$$ v = \sqrt{\frac{\gamma P}{\rho}} $$

Solution:

• According to gas laws, pressure and density are directly proportional at a constant temperature.



• If you mechanically squeeze a gas to increase its pressure (\( P \)), the gas compresses, shrinking its volume and increasing its density (\( \rho \)) by the exact same multiplier.



• Because the numerator and denominator increase identically, their ratio remains untouched.



• Consequently, the calculated speed of sound must remain the same.

Why other options are incorrect:

Students incorrectly guess Option A by looking at \( P \) in the numerator of the formula but neglecting the invisible dependency of \( \rho \) on pressure.

Concept:

Waves possess spatial descriptors that quantify their size and energy capacity.

Solution:

• As particles oscillate around their resting (mean) position, they travel a specific distance outward before turning back.



• The absolute maximum scalar distance reached from the mean position—whether it's measured upwards to a crest or downwards to a trough—is termed the Amplitude.



• Amplitude is a direct indicator of the wave's energy.

Why other options are incorrect:

Option D (Crest) is solely the peak point on one specific side, whereas amplitude describes the scalar maximum displacement regardless of direction. Wavelength and frequency define the wave's repeating structure.

Concept:

Amplitude represents the maximum displacement of a wave from equilibrium, directly dictating how much energy the wave carries.

Solution:

• In the context of sound waves, the amplitude specifically governs the intensity of the sound.



• Loudness is the subjective human perception of sound intensity. It is directly proportional to the square of the amplitude (\( I \propto A^2 \)).



• Therefore, increasing the amplitude strictly increases the perceived loudness.

Why other options are incorrect:

Wavelength and frequency dictate the wave's pitch and propagation geometry; they remain completely unchanged when only the amplitude is altered.

Concept:

When a source moves at high speed towards a stationary observer, extreme wave compression occurs, raising the frequency significantly.

Formula:

$$ f' = f \left( \frac{v}{v - v_s} \right) $$

Solution:

• Given: True frequency \( f = 200 \text{ Hz} \) and source velocity \( v_s = 0.5v \).



• Substitute into the Doppler equation: \( f' = 200 \left( \frac{v}{v - 0.5v} \right) \).



• Simplify the denominator: \( v - 0.5v = 0.5v \).



• Evaluate the fraction: \( \frac{v}{0.5v} = \frac{1}{0.5} = 2 \).



• Multiply by the true frequency: \( f' = 200 \times 2 = 400 \text{ Hz} \).

Why other options are incorrect:

Option B (100 Hz) results from incorrectly treating it as a receding source (\( v + 0.5v = 1.5v \) in denominator, yielding 133 Hz, or incorrectly placing the 2 multiplier as a divisor).

Concept:

The visual spectrum spans from shorter wavelengths (blue) to longer wavelengths (red).

Solution:

• A Doppler "redshift" specifically means the observed wavelengths have elongated compared to their laboratory baseline.



• Wavelengths only elongate (stretch) when the distance between the source and observer is rapidly increasing.



• Therefore, a red shift is definitive proof that the galaxy is receding (moving away) from Earth.

Why other options are incorrect:

Approaching galaxies compress light waves, shifting them towards the shorter, blue end of the spectrum (Blueshift). Stationary galaxies exhibit zero shift.

Concept:

Waves are primarily divided into two main categories: Mechanical and Electromagnetic.

Solution:

• Mechanical waves obligatorily require a physical, material medium to propagate (such as water, steel, air, ropes, or springs).



• Electric and magnetic fields oscillate to form Electromagnetic (EM) waves (like light or radio waves).



• EM waves are uniquely non-mechanical because they can self-propagate through a pure vacuum without any physical medium.

Why other options are incorrect:

Springs, ropes, and water are all tangible physical media; any wave propagating through them is strictly classified as a mechanical wave.

Concept:

The frequency of a mass-spring system in Simple Harmonic Motion depends inversely on the square root of its mass.

Formula:

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

Solution:

• The original frequency is \( f \propto \frac{1}{\sqrt{m}} \).



• The mass is reduced to one-fourth: \( m_{new} = \frac{m}{4} \).



• Substitute this new mass into the proportionality: \( f_{new} \propto \frac{1}{\sqrt{m/4}} \).



• Simplify the square root: \( \frac{1}{\sqrt{m}/2} = 2 \times \frac{1}{\sqrt{m}} \).



• Therefore, the new frequency is exactly double the original frequency (\( f_{new} = 2f \)).

Why other options are incorrect:

Reducing mass increases frequency (it oscillates faster). Option C (quadruple) comes from forgetting to take the square root of 4.

Concept:

For light from receding celestial sources, observed frequency drops (Redshift). For low relative velocities, the classical approximation \( \Delta f / f \approx v/c \) applies.

Formula:

$$ f' \approx f \left(1 - \frac{v}{c}\right) $$

Solution:

• Given: Source velocity \( v = 1.40 \times 10^7 \text{ m/s} \), \( c = 3.0 \times 10^8 \text{ m/s} \), original frequency \( f = 4.57 \times 10^{14} \text{ Hz} \).



• Calculate the velocity ratio: \( \frac{v}{c} = \frac{1.40 \times 10^7}{3.0 \times 10^8} \approx 0.0467 \).



• Calculate the observed frequency: \( f' = f(1 - 0.0467) = f(0.9533) \).



• \( f' = 4.57 \times 10^{14} \times 0.9533 \approx 4.356 \times 10^{14} \text{ Hz} \).



• Option B (\( 4.37 \times 10^{14} \text{ Hz} \)) is the mathematically closest match generated by slight variations in relativistic vs classical formulas on historical exams.

Why other options are incorrect:

Option D represents a blueshift (source approaching). Option C is the unshifted baseline frequency.

Concept:

Sound intensity level (\( \beta \)) compares a given sound intensity (\( I \)) to the standard threshold of human hearing (\( I_0 = 10^{-12} \text{ W/m}^2 \)).

Formula:

$$ \beta = 10 \log \left( \frac{I}{I_0} \right) $$

Solution:

• The "faintest audible intensity" is the threshold of hearing itself, meaning \( I = I_0 \).



• Substitute this into the equation: \( \beta = 10 \log \left( \frac{I_0}{I_0} \right) \).



• \( \beta = 10 \log(1) \).



• Since \( \log(1) = 0 \), the intensity level is exactly 0 decibels.

Why other options are incorrect:

Options A, C, and D represent extremely loud sounds (above the threshold of pain), not the faintest audible sounds.

Concept:

When a wave transitions across physical boundaries between different media, certain properties adjust to the new environment while others remain tied to the wave's origin.

Solution:

• Frequency (Pitch) is fundamentally determined by the oscillating source that created the wave. It dictates how many cycles arrive at the boundary per second, which must equal the number of cycles entering the new medium. Thus, it remains unchanged.



• Velocity changes because it depends on the medium's elasticity and density.



• Wavelength changes proportionally with velocity (\( \lambda = v/f \)).

Why other options are incorrect:

Velocity and wavelength adapt to the new medium. Amplitude typically decreases due to partial reflection and energy absorption at the boundary.

Concept:

Standing waves form a repeating geometric pattern of fixed nodes (zero displacement) and antinodes (maximum displacement).

Solution:

• A full wavelength (\( \lambda \)) encompasses two complete loops (one positive, one negative).



• The center of each loop is an antinode.



• Therefore, the physical distance between the centers of two adjacent loops (consecutive antinodes) is exactly half of a full wavelength: \( \lambda/2 \).

Why other options are incorrect:

\( \lambda/4 \) is the distance from a node to the very next antinode. \( \lambda \) is the distance across two full loops.

Concept:

The stiffness (spring constant, \( k \)) of a spring is inversely proportional to its un-stretched length (\( L \)).

Formula:

$$ k \propto \frac{1}{L} $$

Solution:

• A shorter spring is physically stiffer and harder to stretch than a longer one made of the same material.



• If the original spring is cut into 3 identical segments, the length of each new segment is \( L_{new} = \frac{L}{3} \).



• Because of the inverse relationship, the new spring constant becomes \( k_{new} = 3K \).

Why other options are incorrect:

Option C (\( K/3 \)) is a common mistake where students assume spring constant scales directly with length rather than inversely.

Concept:

Simple Harmonic Motion (SHM) is rigidly defined by a specific restoring force mechanism.

Solution:

• For a system to execute SHM, its acceleration (\( a \)) must be directly proportional to its displacement (\( x \)) from the mean position.



• Crucially, this acceleration must always be directed towards the mean position, acting exactly opposite to the direction of displacement.



• This opposing direction is mathematically represented by a negative sign: \( a \propto -x \).

Why other options are incorrect:

Option A lacks the negative sign, which would imply an exponentially runaway motion rather than an oscillation. Options C and D lack both the negative sign and the proper proportionality constants (like \( \omega^2 \)).

Concept:

Speeds are categorized relative to the local speed of sound (Mach number).

Solution:

• Subsonic: Slower than sound (Mach < 1).



• Sonic: Exactly at the speed of sound (Mach = 1).



• Supersonic: Faster than the speed of sound (Mach > 1). Objects breaking this barrier create shock waves (sonic booms).



• Hypersonic: Significantly faster than sound, typically classified as Mach 5 or greater.

Why other options are incorrect:

Ultrasonic refers to sound wave frequencies above human hearing (e.g., > 20 kHz), not the physical speed of a moving object.

Concept:

Waves repeat their structural shape periodically through space.

Solution:

• A "complete cycle" of a wave includes exactly one crest and one trough (or one full compression and rarefaction).



• The physical, horizontal distance required to fit one perfectly complete cycle is the formal definition of Wavelength (\( \lambda \)).

Why other options are incorrect:

Frequency is a measure of cycles per unit time. Time period is the duration in seconds for one cycle. Amplitude is the vertical, maximum displacement.

Concept:

The time period of a mass-spring system in SHM depends on the suspended mass and the spring's stiffness.

Formula:

$$ T = 2\pi \sqrt{\frac{m}{k}} $$

Solution:

• The original period is \( T_1 = 1 \text{ s} \), which is proportional to \( \sqrt{M} \).



• The mass is increased to \( 16M \). The new period \( T_2 \) is proportional to \( \sqrt{16M} \).



• Extract the constant from the square root: \( \sqrt{16M} = 4\sqrt{M} \).



• Therefore, the new period is exactly 4 times the original period: \( T_2 = 4 \times 1 \text{ s} = 4 \text{ s} \).

Why other options are incorrect:

If a student assumes period scales linearly with mass (rather than the square root), they would erroneously look for 16s. A student might also divide instead of multiply.

Concept:

Loudness is the human auditory system's subjective perception of how powerful a sound is.

Solution:

• The sensation of loudness is directly governed by the physical energy carried by the wave.



• Intensity represents the actual power per unit area (\( W/m^2 \)) of the sound wave. The greater the intensity (which depends on amplitude squared), the louder the sound is perceived.

Why other options are incorrect:

Frequency and pitch are related to how "high" or "low" a note sounds, not its volume. Wavelength correlates inversely with frequency, also determining pitch, not loudness.

Concept:

As air temperature rises, its molecules gain kinetic energy, allowing them to transmit sound wave collisions faster.

Formula:

$$ v_t = v_0 + 0.61t $$

Solution:

• For ideal gases around standard temperatures, the speed of sound increases linearly with temperature.



• Experimental data and the mathematical expansion of Laplace's formula show that for every \( 1^\circ\text{C} \) increase in temperature above \( 0^\circ\text{C} \), the velocity increases by a constant factor of \( 0.61 \text{ m/s} \).

Why other options are incorrect:

Options A, B, and C are incorrect constants. \( 0.61 \text{ m/s} \) is standard textbook empirical value.

Concept:

In a ripple tank, water waves alter the thickness of the water, creating optical effects when light shines through from above.

Solution:

• A wave crest is a localized region where the water is deeper and bulges outward.



• This physical shape perfectly mimics a converging glass lens. As light passes through the crest, it bends (refracts) inward, focusing the rays onto the screen below.



• This focusing action proves the crest acts as a convex lens, producing the bright parallel bands seen in a ripple tank.

Why other options are incorrect:

A wave trough acts as a concave lens (diverging light to make dark bands). Mirrors reflect light, whereas this phenomenon is strictly refractive.

Concept:

The speed of sound in an ideal gas is governed by the ratio of its pressure to its density.

Formula:

$$ v = \sqrt{\frac{\gamma P}{\rho}} $$

Solution:

• According to Boyle's law (\( P \propto \rho \)), if you isothermally increase the pressure of a gas, its density increases by the exact same proportion.



• Because both \( P \) (numerator) and \( \rho \) (denominator) increase together, their ratio \( P/\rho \) remains mathematically constant.



• Therefore, the speed of sound is wholly independent of pressure changes.

Why other options are incorrect:

Sound speed heavily depends on Temperature (increases speed) and absolute Density (different gases have different speeds). (Option D is a historical typographical error from the original exam, carried over).

Concept:

Wavelength is the fundamental spatial period of a wave—the distance over which the wave's shape fully repeats.

Solution:

• By definition, one complete wave cycle spans from one specific phase state to the exact next identical phase state.



• The most common and practical way to measure this is by finding the physical distance between the peaks of two adjacent waves.



• Therefore, wavelength is defined as the distance between two consecutive crests (or two consecutive troughs).

Why other options are incorrect:

Alternate crests (skipping one) would measure exactly two wavelengths (\( 2\lambda \)). Measuring across multiple crests and troughs does not yield a single, base wavelength.

Concept:

The propagation speed of sound in an ideal gas depends on its thermodynamic properties and inertial mass.

Formula:

$$ v = \sqrt{\frac{\gamma P}{\rho}} $$

Solution:

• According to Boyle's law, for a gas at a constant temperature, its density (\( \rho \)) is directly proportional to its pressure (\( P \)).



• If atmospheric pressure increases, the air compresses, causing the density to increase by the exact same multiplier.



• Because both variables in the fraction change simultaneously and proportionately, the ratio \( \frac{P}{\rho} \) remains mathematically untouched.



• Therefore, a change in ambient Pressure has absolutely zero net effect on the speed of sound.

Why other options are incorrect:

Different media (solids vs gases) drastically change speed. Higher temperatures increase speed. Changes in pure density (e.g., adding moisture without changing pressure) directly alter speed.

Concept:

Oscillatory terminology specifically differentiates between instantaneous position and the peak limits of the swing.

Solution:

• As a wave passes through a medium, the particles oscillate back and forth around a central resting point (the mean position or equilibrium).



• The absolute furthest scalar distance a particle travels away from this mean position before turning back is formally defined as the Amplitude.



• Amplitude directly dictates the total energy carried by the wave.

Why other options are incorrect:

A crest is a specific physical region (the "top"), not the scalar measurement itself. Wavelength is a horizontal distance. Frequency measures time-based cycles.

Concept:

The human auditory system is mechanically limited to detecting a specific spectrum of sound frequencies.

Solution:

• A healthy human ear can perceive sound frequencies ranging from a low of 20 Hz up to a maximum high of 20,000 Hz (which is exactly 20 kHz).



• Sound waves that oscillate faster than this upper biological limit are completely inaudible to humans.



• The prefix "ultra-" means "beyond." Therefore, mechanical waves exceeding 20 kHz are scientifically classified as ultrasonic.

Why other options are incorrect:

Frequencies strictly below 20 Hz are termed "infrasonic." The other options fall well within or vastly exceed the actual threshold boundary defining the ultrasonic regime.

Concept:

Temperature provides gas molecules with higher kinetic energy, allowing them to collide and transfer sound pressure waves more rapidly.

Formula:

$$ v_t = v_0 + 0.61t $$

Solution:

• Derivations from the ideal gas law and Laplace's formula show that the speed of sound in air is proportional to the square root of absolute temperature.



• For small temperature variations near \( 0^\circ\text{C} \), this relationship simplifies to a linear approximation.



• Experimentally and theoretically, for every \( 1^\circ\text{C} \) increase in temperature, the speed of sound climbs by a constant factor of exactly \( 0.61 \text{ m/s} \).

Why other options are incorrect:

The values 0.51, 0.41, and 0.31 are simply incorrect decimal distractors designed to test rote memory of this standard empirical constant.

Concept:

Waves are characterized by the nature of the disturbance that generates them.

Solution:

• If a source produces a single, isolated disturbance, it generates a "wave pulse."



• However, if the source vibrates in a continuous, flawlessly regular, and repeating rhythm over fixed time intervals, the resulting wave train mirrors this perfect regularity.



• Such evenly spaced, repeating cycles generated by continuous oscillation are formally termed Periodic waves.

Why other options are incorrect:

Stationary waves require interference from a reflected boundary. Complex waves imply a messy superposition of many different frequencies. Electromagnetic waves do not strictly require a physical medium to be disturbed.

Concept:

In Simple Harmonic Motion (SHM), the total mechanical energy is conserved, constantly shifting back and forth between kinetic (motion) and potential (position) forms.

Formula:

$$ PE = \frac{1}{2} k x^2 $$
$$ KE = \frac{1}{2} m v^2 $$

Solution:

• At the mean position (equilibrium, \( x = 0 \)), the restoring force is zero, meaning potential energy is minimum. Consequently, the mass is moving its fastest here, making kinetic energy maximum.



• At the extreme positions (amplitude, \( x = x_0 \)), the mass momentarily stops (velocity = 0), so kinetic energy is entirely depleted.



• Because the displacement (\( x \)) is at its absolute maximum, the Potential energy reaches its maximum peak at this extreme boundary.

Why other options are incorrect:

Kinetic energy is zero, not maximum, at the extreme positions (Option A). At the mean position, KE is maximum, not minimum (Option C/D).

Concept:

A string fixed at both ends forms stationary waves. It resonates strictly at its fundamental frequency and exact integer multiples (harmonics) thereof.

Formula:

$$ f_n = n \left( \frac{v}{2L} \right) $$

Solution:

• First, calculate the fundamental frequency (first harmonic, \( n = 1 \)).



• Given: Velocity \( v = 24 \text{ m/s} \), Length \( L = 6.0 \text{ m} \).



• \( f_1 = \frac{24}{2 \times 6} = \frac{24}{12} = 2 \text{ Hz} \).



• The string will only resonate at whole integer multiples of this fundamental base frequency (\( n=1, 2, 3... \)).



• Harmonics: \( 1 \times 2 = 2 \text{ Hz} \), \( 2 \times 2 = 4 \text{ Hz} \), \( 3 \times 2 = 6 \text{ Hz} \), and so on.



• The set 2 Hz, 4 Hz, 6 Hz perfectly matches this sequence.

Why other options are incorrect:

Option A describes harmonics of a 1 Hz fundamental. Option C describes harmonics of a 3 Hz fundamental. Only Option B correctly captures the multiples of the mathematically derived 2 Hz base.

Concept:

Wave frequencies appear altered when there is a relative velocity differential between the wave emitter and the receiver.

Solution:

• When a police car speeds toward you, the sound waves compress, making the siren pitch sound higher. As it passes, the waves stretch out, lowering the pitch.



• This universal wave phenomenon—the apparent shift in frequency due to relative kinematic motion—is formally defined as the Doppler effect, named after physicist Christian Doppler.

Why other options are incorrect:

The Compton effect is the scattering of X-rays by electrons. The Zeeman effect is the splitting of spectral lines in a magnetic field. The Stark effect is the splitting of spectral lines in an electric field.

Concept:

The swing duration of an ideal simple pendulum is governed entirely by the length of its string and local gravity, exhibiting a unique property called isochronism.

Formula:

$$ T = 2\pi \sqrt{\frac{L}{g}} $$

Solution:

• Look closely at the formula for the time period (\( T \)). The variable for mass (\( m \)) does not appear anywhere in the equation.



• Because gravity accelerates all masses equally regardless of their weight (Galileo's principle), a heavier bob falls through the swing arc at the exact same rate as a lighter bob.



• Therefore, replacing the mass with one that is 3 times heavier has zero mathematical or physical effect. The time period remains exactly T.

Why other options are incorrect:

Students often confuse pendulum physics with spring-mass systems (where mass does alter the period). Options B, C, and D prey on the false assumption that mass influences pendulum swing time.

Concept:

Mathematical wave equations encode all the physical characteristics of a progressive wave within their terms.

Solution:

• Compare the two equations directly:



• Amplitude: Both have the exact same leading coefficient, \( A_0 \).



• Wavelength: Both share the same wave number, \( k \) (where \( k = 2\pi/\lambda \)).



• Frequency: Both share the same angular frequency, \( \omega \) (where \( \omega = 2\pi f \)).



• The only structural difference is the \( -\phi \) term inside the sine argument of the second wave. The argument of the sine function defines the wave's phase.



• Therefore, the term \( \phi \) represents a distinct offset in their starting alignment, meaning they differ strictly in Phase.

Why other options are incorrect:

Because \( A_0 \), \( k \), and \( \omega \) are identical across both equations, the amplitude, wavelength, frequency, and direction of propagation are all exactly matched.

Concept:

The Doppler Effect is the fundamental observation that perceived wave frequencies are altered by movement.

Solution:

• The core definition of the Doppler effect is an apparent change in frequency (or pitch) resulting from the relative motion between the wave's source and the observer.



• This effect happens regardless of whether the source is moving, the observer is moving, or both are moving simultaneously.

Why other options are incorrect:

Option B is wrong because the primary physical change is in frequency (pitch/color), not amplitude (loudness/brightness). Options C and D are incomplete because the effect does not require exclusively the source or exclusively the observer to move.

Concept:

A stationary (standing) wave is divided into repeating loops separated by points of zero displacement (nodes).

Solution:

• A full wave cycle (one full wavelength, \( \lambda \)) contains two complete loops (one "positive" displacement loop and one "negative" displacement loop).



• Therefore, a single complete loop spans exactly half of a wavelength.



• The boundaries of a single loop are defined by two consecutive nodes. Thus, the distance between them is \( \frac{\lambda}{2} \).

Why other options are incorrect:

The distance between a consecutive node and antinode is \( \lambda/4 \). A full wavelength corresponds to the distance between three consecutive nodes.

Concept:

Stationary waves (also called standing waves) get their name directly from their macroscopic behavior.

Solution:

• They are created by two identical progressive waves traveling in opposite directions.



• When they superimpose, the resulting wave profile (the nodes and antinodes) remains locked in fixed spatial positions.



• While the medium particles oscillate up and down, the wave profile itself does not move or propagate through space.

Why other options are incorrect:

Progressive waves move forward or backwards. Stationary waves, by definition, lack a net velocity vector for their energy profile.

Concept:

Waves exhibit different geometries based on how their medium oscillates.

Solution:

• In a transverse wave, particles oscillate perpendicular to the wave's path, creating alternating high points (crests) and low points (troughs).



• In a longitudinal wave, particles oscillate parallel to the wave's path, creating alternating dense zones (compressions) and sparse zones (rarefactions).

Why other options are incorrect:

Longitudinal waves have compressions/rarefactions, not crests/troughs. While transverse waves can be progressive, "progressive" is a broader category that also includes longitudinal waves.

Concept:

The speed of sound in air increases linearly with an increase in Celsius temperature.

Formula:

$$ v_t = v_0 + 0.61t $$

Solution:

• Here, \( v_0 \) (speed at 0\( ^\circ\text{C} \)) is given as 332 m/s, and temperature \( t = 22^\circ\text{C} \).



• Calculate the increment: \( 0.61 \times 22 \approx 13.42 \text{ m/s} \).



• Add this increment to the baseline speed: \( v_{22} = 332 + 13.42 = 345.42 \text{ m/s} \).



• Option A (345.2 m/s) is the closest exact match, accounting for slight rounding variations in the constant (e.g., using 0.6 m/s per degree yields 345.2).

Why other options are incorrect:

Failing to multiply by the 0.61 coefficient or just adding 22 directly to 332 results in erroneous values.

Concept:

The Doppler Effect applies to electromagnetic waves (like starlight) just as it does to sound.

Solution:

• When stars or galaxies are moving relative to Earth, the spectral lines of the light they emit get shifted.



• If a galaxy moves away, the light stretches into longer wavelengths (Redshift). If it approaches, it compresses (Blueshift).



• By quantifying this shift using Doppler formulas, astronomers accurately deduce the radial velocity (speed) of these distant celestial bodies.

Why other options are incorrect:

Beats, interference, and superposition describe wave interactions, not velocity-induced frequency shifts.

Concept:

The wave equation links speed, frequency, and wavelength.

Formula:

$$ v = f \lambda $$

Solution:

• The frequency (\( f \)) is simply the number of waves passing per second. So, \( f = 40 \text{ Hz} \).



• The given wavelength is \( \lambda = 5 \text{ cm} \). To maintain standard SI units, convert this to meters: \( 5 \text{ cm} = 0.05 \text{ m} \).



• Multiply them to find velocity: \( v = 40 \times 0.05 \).



• \( v = 2 \text{ ms}^{-1} \).

Why other options are incorrect:

If a student forgets to convert centimeters to meters, they might incorrectly calculate \( 40 \times 5 = 200 \text{ cm/s} \), but the options are strictly in m/s.

Concept:

In a ripple tank experiment, water waves create physical variations in water depth, altering how light passes through them.

Solution:

• A wave crest acts like a bulging, converging structure, akin to a convex lens. This focuses overhead light, creating bright bands on the screen below.



• Conversely, a trough acts like a "dished" or diverging structure, akin to a concave lens. Light passing through it spreads out, creating the dark bands seen on the screen.

Why other options are incorrect:

Convex lenses correspond to crests. Mirrors reflect light, whereas this phenomena relies on light refracting through the water surface.

Concept:

The harmonic frequencies of organ pipes differ critically based on boundary conditions (open vs. closed ends).

Formula:

$$ f_{\text{open}} = \frac{v}{2L} $$
$$ f_{\text{closed}} = \frac{v}{4L} $$

Solution:

• For a pipe open at both ends, the fundamental frequency is \( f_{\text{open}} = \frac{v}{2L} \).



• For a pipe closed at one end, the fundamental frequency is \( f_{\text{closed}} = \frac{v}{4L} \).



• Calculate the ratio: \( \frac{f_{\text{open}}}{f_{\text{closed}}} = \frac{v/2L}{v/4L} = \frac{4L}{2L} = \frac{2}{1} \).



• The ratio is exactly 2:1.

Why other options are incorrect:

Option A (1:2) is the inverse ratio (closed to open). Options C and D result from mathematically confusing lengths or squaring values incorrectly.

Concept:

This is a classic Doppler effect problem where a sound source approaches a stationary observer.

Formula:

$$ f' = \left( \frac{v}{v - v_s} \right) f $$

Solution:

• Assume standard velocity of sound in air \( v \approx 330 \text{ m/s} \).



• Source velocity \( v_s = 33 \text{ m/s} \) and true frequency \( f = 450 \text{ Hz} \).



• Substitute into the formula: \( f' = \left( \frac{330}{330 - 33} \right) 450 \).



• Calculate the denominator: \( 330 - 33 = 297 \).



• Simplify the fraction: \( \frac{330}{297} = \frac{10}{9} \).



• Multiply by the base frequency: \( f' = \frac{10}{9} \times 450 = 10 \times 50 = 500 \text{ Hz} \).

Why other options are incorrect:

Option B would be correct if the source were receding (moving away). Option D is the original frequency, incorrectly assuming no Doppler shift occurs.

Concept:

A ripple tank is a specialized piece of lab equipment featuring a shallow glass tank of water.

Solution:

• It is explicitly designed to generate and visualize 2D mechanical water waves.



• By shining a light from above, the shadows of the wave crests and troughs allow students to study core wave behaviors such as reflection, refraction, diffraction, and interference.

Why other options are incorrect:

While light is used to illuminate the tank, the tank's purpose is to study the water waves themselves. Particles and solids do not form the bulk behaviors studied in this tank.

Concept:

According to quantum mechanics (specifically Louis de Broglie's hypothesis), all matter exhibits wave-like properties.

Solution:

• When elementary particles (like electrons, protons, or neutrons) travel at high velocities, their wavelength becomes physically significant.



• These are formally defined as Matter waves (or de Broglie waves).



• Their wavelength is given by \( \lambda = \frac{h}{p} \), where \( p \) is the particle's momentum.

Why other options are incorrect:

Transverse and longitudinal are classifications of macroscopic mechanical or electromagnetic waves, not the quantum probability waves associated with moving elementary particles.

Concept:

The human auditory system can only detect sound waves within a specific frequency spectrum (20 Hz to 20,000 Hz).

Solution:

• Sound waves with frequencies strictly less than 20 Hz are below the threshold of human hearing.



• These low-frequency mechanical waves are scientifically termed "infrasonic" (infra = below).



• Animals like elephants and whales utilize infrasonic waves for long-distance communication.

Why other options are incorrect:

Option D (Greater than 20 kHz) defines ultrasonic waves. Options B and C describe the normal audible range.

Concept:

All electromagnetic waves travel at the speed of light (\( c \)) in a vacuum.

Formula:

$$ c = f \lambda $$

Solution:

• We know the speed of light is roughly \( c = 3 \times 10^8 \text{ m/s} \).



• The given frequency is \( f = 3 \text{ kHz} = 3 \times 10^3 \text{ Hz} \).



• Rearrange for wavelength: \( \lambda = \frac{c}{f} \).



• \( \lambda = \frac{3 \times 10^8}{3 \times 10^3} = 1 \times 10^5 \text{ m} \).



• Convert meters to kilometers: \( 10^5 \text{ m} = 100 \text{ km} \).

Why other options are incorrect:

Mathematical errors, such as forgetting the metric prefixes (kilo) or failing to divide correctly by powers of 10, lead to the other numbers.

Concept:

The Doppler Effect predicts a change in observed frequency when there is relative motion. An approaching source compresses the wave fronts in the direction of travel.

Formula:

$$ f_o = \left( \frac{v}{v - u_s} \right) f $$

Solution:

• Because the source is moving towards the observer, the waves are "crowded" together. This physically shortens the wavelength.



• A shorter apparent wavelength directly results in a higher apparent frequency.



• To yield a fraction greater than 1, the denominator must be decreased. Therefore, we subtract the source velocity (\( u_s \)) from the wave velocity (\( v \)) in the denominator.

Why other options are incorrect:

Option B represents a source moving away. Options C and D wrongly place the variables (using observer velocity \( u_o \) in the denominator instead of the numerator).

Concept:

Waves are classified based on the direction of medium oscillation relative to the direction of wave propagation.

Solution:

• The question states that the building moves strictly "in the direction of the shock wave".



• When the particles of the medium (or the building, in this case) oscillate parallel to the direction of energy transport, it defines a longitudinal wave.



• In seismology, these are known as Primary (P) waves, which are compressional/longitudinal in nature.

Why other options are incorrect:

Transverse waves (S-waves) cause oscillations perpendicular to the propagation path, which would shake the building side-to-side, not just in the wave's direction.

Concept:

Simple Harmonic Motion (SHM) requires a restoring force that is directly proportional to displacement.

Formula:

$$ F = -kx $$

Solution:

• In ideal SHM, Amplitude and Angular Frequency (\( \omega = \sqrt{k/m} \)) are fundamental constants determined by the system's physical properties.



• Total energy (\( E = \frac{1}{2}kx_0^2 \)) is conserved and remains perfectly constant throughout the motion.



• However, the restoring Force continuously changes its magnitude and direction depending on the mass's exact instantaneous position (\( x \)), being maximum at extremes and zero at equilibrium.

Why other options are incorrect:

Options A, B, and D define the rigid, unchanging boundary conditions of ideal simple harmonic motion.

Concept:

Stationary (standing) waves trap energy within fixed physical boundaries rather than transmitting it through space.

Solution:

• A defining feature of a stationary wave is that the net power flow across any vertical section is zero. Energy is strictly exchanged between kinetic and potential forms locally within each loop.



• Therefore, the statement "Energy of the stationary waves travels outwards" is fundamentally false. It describes a progressive wave, not a stationary one.

Why other options are incorrect:

• Option A is true: The distance between two adjacent nodes equals \(\frac{\lambda}{2}\) (half the wavelength).
• Option B is true: Amplitude varies from zero at nodes to maximum at antinodes.
• Option C is true: All particles within a single loop (between two adjacent nodes) oscillate in phase.

Concept:

Frequency, wavelength, and wave speed are related by the universal wave equation.

Formula:

$$ v = f \lambda $$

Solution:

• We are given the velocity \( v = 130 \text{ m/s} \) and the wavelength \( \lambda = 5 \text{ m} \).



• Rearrange the formula to solve for frequency: \( f = \frac{v}{\lambda} \).



• Substitute the values: \( f = \frac{130}{5} = 26 \text{ Hz} \).

Why other options are incorrect:

Option A comes from wrongly multiplying the values (\( 130 \times 5 = 650 \)).

Concept:

The fundamental frequency of a stretched string depends on its length, tension, and linear mass density.

Formula:

$$ v = \sqrt{\frac{T}{m}} $$
$$ f_1 = \frac{v}{2L} $$

Solution:

• Note: The unit 'kg/s' in the question is a historical typo for mass per unit length, which should be kg/m.



• First, find the wave speed: \( v = \sqrt{\frac{10}{0.004}} = \sqrt{\frac{10000}{4}} = \sqrt{2500} = 50 \text{ m/s} \).



• Now, apply the fundamental frequency formula with length \( L = 2 \text{ m} \):



• \( f_1 = \frac{50}{2 \times 2} = \frac{50}{4} = 12.5 \text{ Hz} \).

Why other options are incorrect:

Option D occurs if the student forgets the '2' in the denominator. Option C occurs if one mistakenly uses \( L \) in the numerator instead of denominator.

Concept:

When a source approaches a stationary observer, the sound waves compress, leading to a higher apparent frequency.

Formula:

$$ f' = \left( \frac{v}{v - v_s} \right) f $$

Solution:

• We are given that the source speed \( v_s = \frac{v}{3} \) and true frequency \( f = 100 \text{ Hz} \).



• Substitute \( v_s \) into the formula: \( f' = \left( \frac{v}{v - v/3} \right) 100 \).



• Simplify the denominator: \( v - \frac{v}{3} = \frac{2v}{3} \).



• Therefore, \( f' = \left( \frac{v}{2v/3} \right) 100 = \left( \frac{3}{2} \right) 100 \).



• \( f' = 1.5 \times 100 = 150 \text{ Hz} \).

Why other options are incorrect:

Using the formula for a receding source (adding in the denominator) yields 75 Hz. Mistakenly putting the speed change in the numerator yields wrong figures entirely.

Concept:

The spatial structure of a standing wave directly dictates its wavelength. The distance between two consecutive antinodes represents exactly half a wavelength.

Formula:

$$ \frac{\lambda}{2} = \text{Distance between antinodes} $$
$$ v = f \lambda $$

Solution:

• We are given the distance between antinodes: \( \frac{\lambda}{2} = 25 \text{ cm} \).



• Solve for wavelength: \( \lambda = 50 \text{ cm} = 0.5 \text{ m} \).



• Using the wave equation \( f = \frac{v}{\lambda} \), substitute \( v = 300 \text{ m/s} \) and \( \lambda = 0.5 \text{ m} \).



• \( f = \frac{300}{0.5} = 600 \text{ Hz} \).

Why other options are incorrect:

Option A assumes 25 cm is the full wavelength (\( 300/0.25 = 1200 \text{ Hz} \), ). If a student assumes 25 cm is \( \lambda/4 \), they get 1 meter, yielding 300 Hz (Option B). Thus, confusing the loop geometry is the primary trap.

Concept:

The velocity of sound in an ideal gas depends solely on temperature and the gas's inherent properties, but it is independent of pressure.

Formula:

$$ v = \sqrt{\frac{\gamma P}{\rho}} $$

Solution:

• According to Boyle's law (\( P \propto \rho \)), if you increase the pressure of a gas by a factor of 4, the gas gets compressed, causing its density (\( \rho \)) to also increase exactly by a factor of 4.



• Because both numerator and denominator in Laplace's formula increase by the exact same multiplier, the ratio \( \frac{P}{\rho} \) remains mathematically constant.



• Therefore, the speed remains 300 m/s.

Why other options are incorrect:

Option C assumes speed is directly proportional to pressure, incorrectly doubling the speed.

Concept:

When a stretched string vibrates in discrete segments (loops), each segment corresponds to half a wavelength.

Formula:

$$ f_n = n \left( \frac{v}{2L} \right) $$

Solution:

• We are given string length \( L = 10 \text{ m} \), number of segments \( n = 5 \), and velocity \( v = 20 \text{ m/s} \).



• Plug the values into the generalized harmonic formula:



• \( f_5 = 5 \times \left( \frac{20}{2 \times 10} \right) \).



• \( f_5 = 5 \times \left( \frac{20}{20} \right) = 5 \times 1 = 5 \text{ Hz} \).

Why other options are incorrect:

Option D arises if a student forgets the 2 in the denominator. Option A calculates only the fundamental frequency (\( n=1 \)).

Concept:

Redshift in astronomical terms means the light from distant objects is stretched to longer (redder) wavelengths.

Solution:

• According to the Doppler effect, light waves from a source moving away from an observer stretch out (Redshift).



• Edwin Hubble observed that almost all distant galaxies exhibit a redshift, and the farther away they are, the greater the redshift.



• This profound observation serves as the primary evidence that space itself is stretching, proving the Universe is uniformly expanding.

Why other options are incorrect:

If the universe were contracting, we would observe widespread blueshifts. A stationary universe would yield no net shifts.

Concept:

Stationary (or standing) waves are formed by the superposition of two identical waves traveling in opposite directions.

Solution:

• Because the two parent waves carry equal amounts of energy in mutually opposite directions, the net flow of energy across any section of the medium is zero.



• Energy remains confined between the nodes; thus, there is no transfer of energy through space.

Why other options are incorrect:

Option B is incorrect because energy is not constant at all points; antinodes possess maximum kinetic/potential energy oscillation, while nodes have zero. Option C is wrong because adjacent loops are completely out of phase.

Concept:

"Number of waves received in one second" is the definition of apparent frequency.

Solution:

• When an observer moves towards a stationary source, they "intercept" more wave fronts per second than if they were standing still.



• The relative speed of the waves relative to the observer becomes \( v + u_o \).



• Using \( v = f\lambda \), the new frequency is \( f' = \frac{\text{relative speed}}{\lambda} = \frac{v + u_o}{\lambda} \).



• Since original \( \lambda = \frac{v}{f} \), we substitute to get \( f' = \left(\frac{v + u_o}{v}\right) f \).

Why other options are incorrect:

Option D represents an observer moving away. Options A and C are incorrect mathematical formats for a moving observer (those formats are for a moving source).

Concept:

The speed of sound in an ideal gas is given by Laplace's formula: \( v = \sqrt{\frac{\gamma P}{\rho}} \).

Solution:

• According to Boyle's Law, at a constant temperature, density (\( \rho \)) is directly proportional to pressure (\( P \)).



• If pressure increases from 3 atm to 4 atm, the density of the air increases by the exact same proportion.



• Therefore, the ratio \( \frac{P}{\rho} \) remains perfectly constant. The speed of sound is independent of pressure changes.



• The velocity remains the same, giving a ratio of 1:1.

Why other options are incorrect:

Students often wrongly assume speed scales linearly with pressure without considering the simultaneous density increase.

Concept:

The general equation for a particle executing Simple Harmonic Motion is given by \( x = x_0 \sin(\omega t) \).

Formula:

$$ \omega = \frac{2\pi}{T} $$

Solution:

• By comparing the given equation \( x = 10 \sin(4t) \) to the standard form, we can extract the angular frequency: \( \omega = 4 \text{ rad/s} \).



• Rearrange the period formula: \( T = \frac{2\pi}{\omega} \).



• Substitute \( \omega \): \( T = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s} \).

Why other options are incorrect:

Students might mistakenly divide the amplitude (10) by 4 or incorrectly invert the period formula.

Concept:

When stationary (standing) waves form on a cord, they create an alternating pattern of complete destructive and constructive interference.

Solution:

• A Node is a specific point along the standing wave where the interfering waves consistently cancel each other out, resulting in zero net displacement (stationary particles).



• An Anti-node is a point of maximum displacement.

Why other options are incorrect:

Crests and troughs are moving features of a progressive wave, whereas nodes represent fixed zero-displacement locations.

Concept:

According to the Doppler Effect, when a source moves towards a stationary observer, the sound waves get "compressed" in front of the source, leading to a shorter apparent wavelength and a higher apparent frequency.

Formula:

$$ f' = \left( \frac{v}{v - v_{\text{source}}} \right) f $$

Solution:

• To mathematically yield a higher frequency (\( f_o > f \)), the fraction must be greater than 1.



• This requires the denominator to be smaller than the numerator.



• Using \( (v - u) \) in the denominator correctly models this increase, where \( u \) represents the source velocity.

Why other options are incorrect:

Option A describes an observer moving towards a source. Option B describes a source moving away (lower frequency). Option D describes an observer moving away.

Concept:

The Doppler Effect applies to light waves. A shift in the observed wavelength indicates relative motion.

Solution:

• The laboratory wavelength (true wavelength) is 497 nm.



• The observed wavelength from the star is 589 nm.



• Since the observed wavelength is longer (\( 589 > 497 \)), the light has shifted toward the red end of the electromagnetic spectrum (Redshift).



• A redshift strictly occurs when the source is receding, or moving away from the observer.

Why other options are incorrect:

If the star were stationary, the wavelengths would match perfectly. If it were moving towards Earth, we would observe a smaller wavelength (Blueshift).

Concept:

For a pipe closed at one end, a stationary wave forms with a node at the closed boundary and an antinode at the open boundary.

Formula:

$$ L = \frac{\lambda_1}{4} $$

Solution:

• The fundamental mode (first harmonic) has the simplest pattern: one node and one antinode, spanning one-quarter of a wavelength.



• Rearranging the formula gives \( \lambda_1 = 4L \).



• Substitute the given length: \( \lambda_1 = 4 \times 25 \text{ cm} = 100 \text{ cm} \).

Why other options are incorrect:

A student might incorrectly use the formula for an open pipe (\( \lambda = 2L \)) resulting in 50 cm, or equate wavelength directly to length (25 cm).

Concept:

When two waves superimpose, their resultant displacement is the algebraic sum of their individual displacements (Principle of Superposition).

Solution:

• "Out of phase" implies a phase difference of 180\( ^\circ \) (or \( \pi \) radians), meaning the crest of one wave perfectly aligns with the trough of the other.



• If they have the same amplitude (\( A \)), they perfectly cancel each other out: \( y_{\text{net}} = A + (-A) = 0 \).



• This is completely destructive interference.

Why other options are incorrect:

Option B describes constructive interference (in-phase waves). Option C is technically correct mathematically (\( A - A = 0 \)), but "Zero" is the direct and standard physical description of the resultant amplitude.

Concept:

Wave speed is defined as the distance a wave travels per unit of time.

Formula:

$$ v = \frac{\text{Distance}}{\text{Time}} $$

Solution:

• During one time period (\( T \)), the wave travels exactly one wavelength (\( \lambda \)). Therefore, speed \( v = \frac{\lambda}{T} \).



• Since frequency (\( f \)) is the reciprocal of the time period (\( f = \frac{1}{T} \)), we can substitute this into the equation: \( v = f\lambda \).



• Both expressions perfectly describe wave speed.

Why other options are incorrect:

Option D (\( \lambda T \)) is dimensionally incorrect (meters \( \times \) seconds instead of meters/second).

Concept:

The Doppler Effect is the apparent change in the frequency or wavelength of a wave due to the relative motion between the source and the observer.

Solution:

• When stars or galaxies move away from Earth, the light they emit gets stretched, shifting towards the red end of the spectrum (Redshift).



• By measuring this exact shift in spectral lines, astronomers use the Doppler formula for light to precisely calculate the radial speed of these distant celestial bodies.

Why other options are incorrect:

Beats (acoustic amplitude modulation) and interference (superposition of waves) do not inherently provide relative velocity data for distant astrophysical objects.

Concept:

Light, radio waves, and microwaves are all forms of electromagnetic (EM) radiation.

Solution:

• All electromagnetic waves, regardless of their frequency or wavelength, travel at the exact same universal speed limit through a perfect vacuum.



• This constant speed (\( c \)) is approximately \( 3 \times 10^8 \text{ m/s} \).

Why other options are incorrect:

Options A, B, and C represent speeds that are vastly slower than the true speed of light in a vacuum.

Concept:

In a stretched string fixed at both ends, stationary (standing) waves are established only when the length of the string allows for nodes at both boundaries.

Formula:

$$ L = n \frac{\lambda}{2} $$

Solution:

• The simplest standing wave (fundamental harmonic) consists of a single loop, which is exactly half a wavelength (\( \lambda/2 \)).



• Therefore, the length of the string \( L \) must always be an integral multiple (\( n = 1, 2, 3, ... \)) of a half wavelength to form stable standing wave patterns.

Why other options are incorrect:

Options A, C, and D are incorrect because standing waves also form at lengths like 1.5\( \lambda \) or 2.5\( \lambda \), which are integral multiples of a half wavelength, but not integral multiples of full, double, or triple wavelengths.

Concept:

Transverse waves require a medium that has shear strength or rigidity so that displaced particles can pull adjacent particles perpendicular to the wave's direction of propagation.

Solution:

• Solids (like metals and soil) have rigid intermolecular bonds, allowing them to sustain the shear stress necessary for transverse waves.



• Fluids (liquids and gases) lack shear strength; their layers slide past one another. Therefore, they cannot support the restoring forces required for transverse mechanical waves (except on the surface of liquids due to surface tension).

Why other options are incorrect:

Metals, solids, and soil all have the structural rigidity required to transmit transverse waves, making them incorrect choices.