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When a mass of 0.5 kg is suspended from ...

When a mass of 0.5 kg is suspended from the free end of a spring, it stretches the spring by 0.2m. This mass is removed and 0.25kg mass is attached to the same free end of the spring. If the mass is pulled down and released, what is its time period? (`g= 10 ms^(-2)`)

Text Solution

Verified by Experts

When 0.3 kg mass is suspended, extension of the spring is 0.2m.
Since in equilibrium mg `=kx`(or `0.5 xx 10 = k xx 0.2 implies k = 25` N/m
The time period of spring block system is `T= 2pi sqrt(m/k) = 2pi sqrt((0.25)/(25)) = 0.63 s `
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