Two identical tuning forks vibrating at a common frequency of 256 Hz are held stationary at some distance apart. A listener runs between the tuning forks at a speed of 3.0 m/s , approaching one tuning fork while

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Quantrex Academy · Accepted Answer

Correct answer: D. 4.6 Hz

The apparent frequency heard by the listener from the tuning fork being approached is given by Doppler's effect:

f_1 = f \left( \dfrac{v + v_L}{v} \right)

The apparent frequency heard from the tuning fork being receded from is:

f_2 = f \left( \dfrac{v - v_L}{v} \right)

The beat frequency observed by the listener is the difference between these two frequencies:

f_b = f_1 - f_2 = f \left( \dfrac{v + v_L}{v} \right) - f \left( \dfrac{v - v_L}{v} \right)

f_b = f \left( \dfrac{2v_L}{v} \right)

Substituting the given values (

f = 256 \text{ Hz}

,

v_L = 3.0 \text{ m/s}

,

v = 332 \text{ m/s}

):

f_b = 256 \times \left( \dfrac{2 \times 3.0}{332} \right)

f_b = \dfrac{1536}{332} \approx 4.6 \text{ Hz}

Two identical tuning forks vibrating at a common frequency of 256...

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