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(A) The observer hears two sounds: one directly from the source and the other from the reflected sound (echo) from the wall.For the direct sound,the s

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$7.8$

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(A) The observer hears two sounds: one directly from the source and the other from the reflected sound (echo) from the wall.For the direct sound,the source is moving away from the observer. The apparent frequency $n_1$ is given by:$n_1 = n \left( \frac{v}{v + v_S} ight) = 256 \left( \frac{330}{330 + 5} ight) = 256 \left( \frac{330}{335} ight) \approx 252.18 \ Hz$.For the reflected sound,the wall acts as a stationary source reflecting the sound. The source is moving towards the wall,so the wall receives sound at a higher frequency,which it then reflects back to the observer. The apparent frequency $n_2$ is given by:$n_2 = n \left( \frac{v}{v - v_S} ight) = 256 \left( \frac{330}{330 - 5} ight) = 256 \left( \frac{330}{325} ight) \approx 260.00 \ Hz$.The number of beats per second is the difference between these two frequencies:$f_{beat} = n_2 - n_1 = n \left( \frac{v}{v - v_S} - \frac{v}{v + v_S} ight) = n \left( \frac{v(v + v_S) - v(v - v_S)}{v^2 - v_S^2} ight) = \frac{2nvv_S}{v^2 - v_S^2}$.Substituting the values:$f_{beat} = \frac{2 	imes 256 	imes 330 	imes 5}{330^2 - 5^2} = \frac{844800}{108900 - 25} = \frac{844800}{108875} \approx 7.76 \ Hz \approx 7.8 \ Hz$.

$A$ source of sound of frequency $256 \ Hz$ is moving rapidly towards a wall with a velocity of $5 \ m/s$. The speed of sound is $330 \ m/s$. If the observer is between the wall and the source,then the number of beats per second heard will be .... $Hz$.

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