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(C) The apparent frequency $n_1$ of the train coming towards the observer is given by $n_1 = n \left( \frac{v}{v - u} \right)$.The apparent frequency

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The apparent frequency $n_1$ of the train coming towards the observer is given by $n_1 = n \left( \frac{v}{v - u} ight)$.The apparent frequency $n_2$ of the train going away from the observer is given by $n_2 = n \left( \frac{v}{v + u} ight)$.The number of beats produced is the difference between the two apparent frequencies: $\Delta n = n_1 - n_2$.$\Delta n = n \left( \frac{v}{v - u} - \frac{v}{v + u} ight) = n \left( \frac{v(v + u) - v(v - u)}{v^2 - u^2} ight) = n \left( \frac{2uv}{v^2 - u^2} ight)$.Substituting the given values: $n = 300 \; Hz$,$v = 332 \; m/s$,$u = 4 \; m/s$.$\Delta n = 300 	imes \left( \frac{2 	imes 4 	imes 332}{332^2 - 4^2} ight) = 300 	imes \left( \frac{2656}{110224 - 16} ight) = 300 	imes \left( \frac{2656}{110208} ight) \approx 300 	imes 0.0241 = 7.23$.Rounding to the nearest integer,the number of beats is $7$.

Two trains,one coming towards and another going away from an observer,both at $4 \; m/s$,produce a whistle simultaneously of frequency $300 \; Hz$. Find the number of beats produced. (Assume the speed of sound $v = 332 \; m/s$)

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