A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears beats with freque

Question

A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears beats with frequency 2.0 Hz. Find the velocity of each tuning fork if their oscillation frequency is $n = 680$ Hz and the velocity of sound in air is 340 m/s.

Answer:

0.50

Solution:

$Δ f = \frac{2 n uv}{v ^{2} - u ^{2}} \approx \frac{2 n u}{v}$ . $2 = \frac{2 \times 680 u}{340} = 4 u$ , so $u = 0.5$ m/s.

TestHub · Accepted Answer

Δ f = v 2 − u 2 2 n uv ​ ≈ v 2 n u ​ . 2 = 340 2 × 680 u ​ = 4 u , so u = 0.5 m/s.

Physics - Sound Wave Question with Solution | TestHub

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