SATHEE: Wave Motion 4 Question 30

Wave Motion 4 Question 30

28. Two tuning forks with natural frequencies of $340 H z$ each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed. The observer hears beats of frequency $3 H z$ . Find the speed of the tuning fork. Speed of sound $= 340 m / s$ .

(1986, 8M)

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Answer:

Correct Answer: 28. (a, b)

Solution:

Given, $f_{1} - f_{2} = 3 H z$

$\begin{aligned} or & f (\frac{v}{v - v_{s}}) - f (\frac{v}{v + v_{s}}) & = 3 \\ or & 340 [\frac{340}{340 - v_{s}}] - 340 [\frac{340}{340 + v_{s}}] & = 3 \\ or & 340 [{(1 - \frac{v_{s}}{340})}^{- 1}] - 340 [{(1 + \frac{v_{s}}{340})}^{- 1}] & = 3 \\ As & v_{s} « 340 m / s \end{aligned}$

Using binomial expansion, we have

$\begin{aligned} 340 (1 + \frac{v_{s}}{340}) - 340 (1 - \frac{v_{s}}{340}) & = 3 \\ ∴ \frac{2 \times 340 \times v_{s}}{340} & = 3 \\ v_{s} & = 1.5 m / s \end{aligned}$

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Wave Motion 4 Question 30

28. Two tuning forks with natural frequencies of 340Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed. The observer hears beats of frequency 3Hz. Find the speed of the tuning fork. Speed of sound =340m/s.

Answer:

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