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Question

In the arrangement shown in figure, the pulleys are small and light and the springs are ideal. K1=25π2 N/m,K2=2K1,K3=4K1 and K4=4K1 are the force constants of the springs. Calculate the period of small vertical oscillations of the block of mass m=2 kg.


A
12 s
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B
45 s
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C
85 s
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D
58 s
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Solution

The correct option is C 85 s
In static equilibrium of the block, tension in the string is exactly equal to its weight.

Let a vertically downward force F be applied on the block to pull it downwards. Equilibrium is again restored when tension in the string is increased by the same amount F. Hence, total tension in the string becomes equal to (mg+F).

Strings are further elongated due to extra tension F in the strings; tension in each spring increases by 2F. Hence, increase in elongation of the springs is 2FK1,2FK2,2FK3 and 2FK4, respectively.

According to the geometry of the arrangement, downward displacement of the block from its equilibrium positon is

y=2(2FK1+2FK2+2FK3+2FK4)...(1)

If the block is released now, it starts to accelerate upwards due to extra tension F in the string. It means, the restoring force on the block is equal to F. From equation (1), acceleration

a=Fm=y4m(1K1+1K2+1K3+1K4)

Since the acceleration of the block is restoring and is directly proportional to displacement y, the block performs SHM.

Time period of oscillations, T=ya

T=2π4m(1K1+1K2+1K3+1K4)

T=4πm(1K1+1K2+1K3+1K4)

Substituing the values, we get

T=4π2(125π2+12×25π2+14×25π2+14×25π2)

T=85 s

Hence, option (c) is the correct answer.

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