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Question
A small block oscillates back and forth on a smooth concave surface of radius R in Figure. Find the time period of small oscillation.
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Solution
It is given that R is the radius of the concave surface.
Let N be the normal reaction force.
Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = g sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.
\[\therefore \theta = \frac{x}{R}\]
\[ \Rightarrow a = g\theta = g\left( \frac{x}{R} \right)\]
\[ \Rightarrow \left( \frac{a}{x} \right) = \left( \frac{g}{R} \right)\]
\[\Rightarrow a = x\frac{g}{R}\]
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.
Time period \[\left( T \right)\] is given by,
\[T = 2\pi\sqrt{\frac{\text { displacement }}{\text { Acceleration }}}\]
\[= 2\pi\sqrt{\frac{x}{gx/R}} = 2\pi\sqrt{\frac{R}{g}}\]
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