A person feels a 2.5% difference in the frequ | vedclass

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(A) According to the Doppler effect,when the source moves towards a stationary observer,the observed frequency $n'$ is given by $n' = n \left( \frac{v

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$8\, m/s$ (approx.)

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(A) According to the Doppler effect,when the source moves towards a stationary observer,the observed frequency $n'$ is given by $n' = n \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound and $v_s$ is the speed of the source.Given the frequency difference is $2.5\%$,we have $\frac{n' - n}{n} = 0.025$,which implies $\frac{n'}{n} = 1.025$.Substituting this into the Doppler formula: $1.025 = \frac{v}{v - v_s}$.Given $v = 320\, m/s$,we have $1.025 = \frac{320}{320 - v_s}$.$320 - v_s = \frac{320}{1.025} \approx 312.195$.$v_s = 320 - 312.195 = 7.805\, m/s$.Rounding to the nearest integer,the velocity of the car is approximately $8\, m/s$.

$A$ person feels a $2.5\%$ difference in the frequency of a motor-car horn. If the motor-car is moving towards the person and the velocity of sound is $320\, m/s$,then the velocity of the car will be:

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