A source of sound attached to the bob of a simple pendulum execute SHM . The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean frequency of the source . The velocity of the source at the mean position is ( velocity of sound in the air is 340 m//s) [Assume velocity of sound lt lt velocity of sound in air ]

A source of sound attached to the bob of a simple pendulum execute SHM

1.	A source of sound attached to the bob of a simple pendulum execute SHM . The difference between the apparent frequency of sound as received by an observer during its approach and recession at the mean frequency of the source . The velocity of the source at the mean position is ( velocity of sound in the air is 340 m//s) [Assume velocity of sound lt lt velocity of sound in air ]
Answer» <html><body><p>`1.4 m//s`<br/>`3.4 m//s`<br/>`1.7 m//s`<br/>`2.1 m//s`</p>Solution :` f_(1) = f_(0) ((V_(0))/(V_(0) - V)) f_(2) = f_(0)((V_(0))/(V_(0) + V))` <br/> `f_(1) - f_(2) = f_(0) V_(0) ((1)/( V_(0) - V) - (1)/( V_(0) + V))` <br/> ` = f_(0) V_(0) (V_(0) + V - V_(0) + V)/( V_(0)^(2) - V^(2)) = f_(0) V_(0) xx ( <a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a>)/( V_(0)^(2)) = f_(0) ( 2V)/( V_(0))` <br/> given `( 2 Vf_(0))/(V_(0)) = 0.02 xx f_(0) rArr V = 0.01 V_(0) = 3.4 m//s`.</body></html>