Looking for classes? Ksquare Career Institute, Bengaluru →
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
$2\pi\sqrt{\frac{x_1^2 + x_2^2}{V_1^2 + V_2^2}}$
B
$2\pi\sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$
C
$2\pi\sqrt{\frac{V_1^2 + V_2^2}{x_1^2 + x_2^2}}$
D
$2\pi\sqrt{\frac{V_1^2 - V_2^2}{x_1^2 - x_2^2}}$
Detailed Solution
In SHM, $v = \omega\sqrt{a^2 - x^2}$.
$V_1^2 = \omega^2(a^2 - x_1^2)$ ...(1)
$V_2^2 = \omega^2(a^2 - x_2^2)$ ...(2)
Subtracting: $V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2) \Rightarrow \omega = \sqrt{\frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}}$
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$
$V_1^2 = \omega^2(a^2 - x_1^2)$ ...(1)
$V_2^2 = \omega^2(a^2 - x_2^2)$ ...(2)
Subtracting: $V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2) \Rightarrow \omega = \sqrt{\frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}}$
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$