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Question

# In the arrangement shown in figure, pulleys are small and light and the springs are ideal. K1=25π2Nm,K2=2K1,K3=3K1 and K4=4K1 are the force constants of the springs. Calculate the period of small vertical oscillations (in s) of the block of mass m=3 kg

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Solution

## 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 the extra tension F. Due to this extra tension F in the strings, tension in each spring increases by 2F [since pulleys are light and massless, force in spring =2× Tension in string. Hence, increase in elongation of the springs are 2FK1,2FK2,2FK3,2FK4 respectively. Accoding to the geometry of the arrangement, downward displacement of the block from its equilibrium position is y=2(2FK1+2FK2+2FK3+2FK4)........(i) If the block is released now, it starts to accelerate upwards due to extra tension F in the strings. It means restoring force on the block is equal to F. From Eq. (I), F=y4(1K1+1K2+1K3+1K4) Restoring acceleration of the blocks a=Fm=y4m(1K1+1K2+1K3+1K4) Since acceleration of the block is resotring and is directly proportional to displacment y, the block performs SHM. Time period, T=2π√displacementacceleration T=2π√4m(1K1+1K2+1K3+1K4) T=4π√m(1K1+1K2+1K3+1K4) T=4π√m×2512K1 After substituting the values, we get T=2 s.

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