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.
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