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Velocity in SHM
Concepts tested here
- v = omega sqrt(a^2 - x^2)
All Questions
2015 AIPMT-I 1 question
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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 isIn 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: $V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2) \Rightarrow \omega = \sqrt{\frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}}$
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$
