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A mass M is suspended from a spring of n...

`A` mass `M` is suspended from a spring of negligible mass. The spring is pulled a little then released, so that the mass executes simple harmonic motion of time period `T`. If the mass is increased by `m`, the time period becomes `(5T)/(3)`. Find the ratio of `m//M`.

Text Solution

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The correct Answer is:
A
`T = 2pi sqrt((M)/(k))`
`(5T)/(3) = 2pi sqrt ((M + m)/(k))`
Solving these two equation, we get
`(m)/(M) = (16)/(9)`
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