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If a particle slides back and forth between two smooth planes as shown in figure then correct statement is

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Te motion is oscillatory

The motion is SHM

If h is the initial height, then its period

T = \frac{4}{\sin θ} \sqrt{\frac{2 h}{g}}

If h is the height, then its period

T = \frac{2}{\sin θ} \sqrt{\frac{2 h}{g}}

answer is C.

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Detailed Solution

T = 4 t

$S = ut + \frac{1}{2} {at}^{2}$

$\frac{h}{\sin θ} = 0 + \frac{1}{2} g \sin θ t^{2}$

$t = \frac{1}{\sin θ} \sqrt{\frac{2h}{g}}$

$T = 4 t = \frac{4}{\sin θ} \sqrt{\frac{2h}{g}}$

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