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A particle is executing SHM along a straight line. Its velocities at distance x_(1)" and "x_(2) form the mean position are v_(1)" and "v_(2), respectively. Its time period is………..

Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/2pi-1838601" style="font-weight:bold;" target="_blank" title="Click to know more about 2PI">2PI</a> <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`<br/>`2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))`<br/>`2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`<br/>`2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`</p>Solution :For <a href="https://interviewquestions.tuteehub.com/tag/shm-630759" style="font-weight:bold;" target="_blank" title="Click to know more about SHM">SHM</a> particle, <br/> `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>= omega sqrt(A^(2)-x^(2))` <br/> `therefore v_(1)= omega sqrt(A^(2)-x_(1)^(2))" and "v_(2)= omega sqrt((A^(2)-x_(2)^(2))` <br/> `therefore v_(1)^(2)= omega^(2)(A^(2)-x_(1)^(2))" and "v_(2)^(2)= omega^(2) (A^(2)-x_(2)^(2))` <br/> `therefore v_(1)^(2) -v_(2)^(2) = omega^(2)[A^(2)-x_(1)^(2)-A^(2)+x_(2)^(2)]` <br/> `= omega^(2) [x_(2)^(2)-x_(2)^(2)]` <br/> `therefore omega^(2)= (v_(1)^(2)-v_(2)^(2))/((x_(2)^(2)-x_(1)^(2)))` <br/> `therefore omega = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` <br/> `therefore (2pi)/(T) = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` <br/> `therefore T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`.</body></html>


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