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Question

A small block oscillates back and forth on a smooth concave surface of radius R. Find the time period of small oscillations.

A
π2Rg
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B
2πR3g
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C
2π2R3g
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D
2πRg
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Solution

The correct option is C 2πRg
Let,
R be the radius of smooth concave surface and m is mass of the block. Also, let the θ is the angle traced by the position of block on smooth concave surface with the normal at mean position while performing small oscillations . So, the displacement, say x of block is,
x=Rsinθ
For small angles sinθ=θ. hence, displacement will be
x=Rθ ...................(1)
The force F responsible for to and fro motion of block is,
F=mgsinθ
F=mgθ .......................(2)
Therefore angular acceleration is,
a=Fm
a=mgθm ...................from(1) and (2)
a=gθ .........................(3)
Now, the angular velocity ω of the block is,
ω=ax
ω=gθRθ
ω=gR ..........................(4)
Put, ω=2πT in (4)
where T is the time period of small oscillations.
2πT=gR
Therefore,
T=2πRg

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