A source (S) of sound has a frequency of 240 | vedclass

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(A) The frequency $f_0$ received by an observer is given by the Doppler effect formula: $f_0 = \left( \frac{C \pm V_0}{C \mp V_s} \right) f_s$,where $

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The frequency $f_0$ received by an observer is given by the Doppler effect formula: $f_0 = \left( \frac{C \pm V_0}{C \mp V_s} ight) f_s$,where $C$ is the speed of sound,$V_0$ is the speed of the observer,and $V_s$ is the speed of the source.Case $1$: Source and observer move towards each other.$f_1 = \left( \frac{C + v}{C - v} ight) f_s$$288 = \left( \frac{C + v}{C - v} ight) 240$$\frac{C + v}{C - v} = \frac{288}{240} = \frac{6}{5} \quad \dots (1)$Case $2$: Source and observer move away from each other.$f_2 = n = \left( \frac{C - v}{C + v} ight) f_s$$n = \left( \frac{C - v}{C + v} ight) 240 \quad \dots (2)$From equation $(1)$,we have $\frac{C - v}{C + v} = \frac{5}{6}$.Substituting this into equation $(2)$:$n = \left( \frac{5}{6} ight) 240$$n = 5 	imes 40 = 200 \ Hz$.

Similar Questions

An observer is moving towards a stationary source of sound. In this case:

$A$ railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside the railway track. The frequency $(n)$ of the sound heard by the observer is plotted against time $(t).$ Which of the following best represents the resulting curve?

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