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In the arrangement shown if Fig. Pulleys...


In the arrangement shown if Fig. Pulleys are small and lught and spring are ideal and `K_1=25(pi^2)(N)/(m)`, `K_2=2K_1`,`K_3=` and `K_4=4K_1` are the force constant of the spring. Calculate the period of small vertical oscillation of block of mass `m=3kg`.

Text Solution

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In static equilibrium of 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 strings, tension in each spring increases by 2F. Hence increase in elongation of springs is `(2F)/(K_1)(2F)/(K_2)(2F)/(K_3)` and `(2F)/(K_4)`, respectively. According to geometry of the arrangement, downward displacement of the block from its equilibrium position is
`y=2((2F)/(K_1)+(2F)/(K_2)+(2F)/(K_3)+(2F)/(K_4))` ..(i)
If the block is released now, it starts to accelerate upeards due to extra tension `F` in the string. It means restoring force on the block is equal to `F`. From Eq. (i)
`F=(y)/(4m((1)/(K_1)+(1)/(K_2)+(1)/(K_3)+(1)/(K_4))`
Since acceleration of block is restoring and is directly proportional to displacement `y`, the block performs `SHM`.
Its period `T=2pisqrt(("displacement")/("acceleration"))`.
`T=2pisqrt(4m((1)/(K_1)+(1)/(K_2)+(1)/(K_3)+(1)/(K_4))`
`T=4pisqrt(m((1)/(K_1)+(1)/(K_2)+(1)/(K_3)+(1)/(K_4))`
After subsituting the values we get `T=2s`.
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