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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 and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?

Text Solution

Verified by Experts

With mass M, the time period of the spring is `T=2pisqrt((M)/(k))`
With mass `M+m`, the time period becomes
`(5T)/(3)=2pisqrt((M+m)/(k))or(5)/(3)xx2pisqrt((M)/(k))=2pisqrt((M+m)/(k))`
`(25)/(9)M=M+m,(16)/(9)M=mor(m)/(M)=(16)/(9)`
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