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In the arrangement shown in figure, pulleys are light and springs are ideal with equal spring constant of 100 N/m. The time period of small vertical oscillations of block of mass 1 kg is


A
0.2π s
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B
0.4π s
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C
0.8π s
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D
0.6π s
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Solution

The correct option is C 0.8π s
FBD of the sytem:


If block will be displaced down by x then each spring will extend by restoring force developed.

Let us suppose, restoring force is F, then extension in each spring will be 2FK.

So, the sum of extensions will be equal to the displacement of block x.

2(2FK)+2(2FK)+2(2FK)+2(2FK)=x

16FK=x

F=Kx16

Restoring force in the spring is equal to the force acting on the block.

ma=Kx16
where, a is acceleration of the block in downward direction.

Substituting the given values,

1×a=100×x16

xa=425

Now, time period,

T=2πxa

T=2π425

T=4π5=0.8π s

Hence, option (c) is correct answer.

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