A particle is moving with shm in a straight line. When the distance of the particle from the equilibrium position has the value x_1 and x_2the corresponding values of velocities are v_1and v_2show that period is T=2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)

A particle is moving with shm in a straight line. When the distance of

1.	A particle is moving with shm in a straight line. When the distance of the particle from the equilibrium position has the value x_1 and x_2the corresponding values of velocities are v_1and v_2show that period is T=2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> at a <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> `x_1` is `v_1=omegasqrt(a^2-x_1^2)` <br/> and `v_2=<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</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>(a^2-x_2^2)` <br/><a href="https://interviewquestions.tuteehub.com/tag/squaring-3058178" style="font-weight:bold;" target="_blank" title="Click to know more about SQUARING">SQUARING</a> and subtracting <br/> `v_1^2-v_2^2=omega^2[a^2-x_1^2-a^2+x_2^2]` <br/> `omega^2=(v_1^2-v_2^2)/(x_2^2-x_1^2)` <br/>`omega=[(v_1^2-v_2^2)/(x_2^2-x_1^2)]^(1//2)` <br/> `(2pi)/T=omega"or " T=(2pi)/omega` <br/>`T = 2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)`</body></html>