Question
A particle is executing $$SHM$$ along a straight line. Its velocities at distances $${x_1}$$ and $${x_2}$$ from the mean position are $${V_1}$$ and $${V_2},$$ respectively. Its time period is
A.
$$2\pi \sqrt {\frac{{x_2^2 - x_1^2}}{{V_1^2 - V_2^2}}} $$
B.
$$2\pi \sqrt {\frac{{V_1^2 + V_2^2}}{{x_1^2 + x_2^2}}} $$
C.
$$2\pi \sqrt {\frac{{V_1^2 - V_2^2}}{{x_1^2 - x_2^2}}} $$
D.
$$2\pi \sqrt {\frac{{x_1^2 - x_2^2}}{{V_1^2 - V_2^2}}} $$
Answer :
$$2\pi \sqrt {\frac{{x_2^2 - x_1^2}}{{V_1^2 - V_2^2}}} $$
Solution :
As we know, for particle undergoing $$SHM,$$
$$\eqalign{
& V = \root \omega \of {{A^2} - {X^2}} ;\,\,V_1^2 = {\omega ^2}\left( {{A^2} - x_1^2} \right) \cr
& V_2^2 = {\omega ^2}\left( {{A^2} - x_2^2} \right) \cr} $$
Substracting we get,
$$\eqalign{
& \frac{{V_1^2}}{{{\omega ^2}}} + x_1^2 = \frac{{V_2^2}}{{{\omega ^2}}} + x_2^2 \cr
& w = \sqrt {\frac{{V_1^2 - V_2^2}}{{x_2^2 - x_1^2}}} \Rightarrow T = 2\pi \sqrt {\frac{{x_2^2 - x_1^2}}{{V_1^2 - V_2^2}}} \cr} $$