Wave Motion Ques 97
- Two tuning forks with natural frequencies of $340 Hz$ 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 $3 Hz$. Find the speed of the tuning fork. Speed of sound $=340 m / s$.
(1986, 8M)
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Answer:
Correct Answer: 97.1.5 m / s
Solution:
Formula:
- Given, $f _1-f _2=3 Hz$
$$ \begin{array}{rlrl} \text { or } & f\left(\frac{v}{v-v _s}\right)-f\left(\frac{v}{v+v _s}\right) & =3 \\ \text { or } & 340\left[\frac{340}{340-v _s}\right]-340\left[\frac{340}{340+v _s}\right] & =3 \\ \text { or } & 340\left[\left(1-\frac{v _s}{340}\right)^{-1}\right]-340\left[\left(1+\frac{v _s}{340}\right)^{-1}\right] & =3 \\ \text { As } & v _s«340 m / s \end{array} $$
Using binomial expansion, we have
$$ \begin{array}{rlrl} 340\left(1+\frac{v _s}{340}\right)-340\left(1-\frac{v _s}{340}\right) & =3 \\ & \therefore \quad \frac{2 \times 340 \times v _s}{340} & =3 \\ v _s & =1.5 m / s \end{array} $$
BETA
