Question
A small block oscillates back and forth on a smooth concave surface of radius R figure. Find the time period of small oscillation.
✓
Answer
Given that, R = radius. Let N = normal reaction. Driving force $\text{F}=\text{mg}\sin\theta.$ Acceleration $=\text{a}=\text{g}\sin\theta$ As, $\sin\theta$ is very small, $\sin\theta\rightarrow\theta$$\therefore$ Acceleration $\text{a}=\text{g}\theta$
Let ‘x’ be the displacement from the mean position of the body,$\therefore\theta=\frac{\text{x}}{\text{R}}$
$\Rightarrow\text{a}=\text{g}\theta=\text{g}\Big(\frac{\text{x}}{\text{R}}\Big)\Rightarrow\Big(\frac{\text{a}}{\text{x}}\Big)=\Big(\frac{\text{g}}{\text{R}}\Big)$
So the body makes S.H.M.$\therefore\text{T}=2\pi\sqrt{\frac{\text{Displacement}}{\text{Acceleration}}}=2\pi\sqrt{\frac{\text{x}}{\frac{\text{gx}}{\text{R}}}}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}$
Let ‘x’ be the displacement from the mean position of the body,$\therefore\theta=\frac{\text{x}}{\text{R}}$
$\Rightarrow\text{a}=\text{g}\theta=\text{g}\Big(\frac{\text{x}}{\text{R}}\Big)\Rightarrow\Big(\frac{\text{a}}{\text{x}}\Big)=\Big(\frac{\text{g}}{\text{R}}\Big)$
So the body makes S.H.M.$\therefore\text{T}=2\pi\sqrt{\frac{\text{Displacement}}{\text{Acceleration}}}=2\pi\sqrt{\frac{\text{x}}{\frac{\text{gx}}{\text{R}}}}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}$
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