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Q.

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 become 5T/3. Then the ratio ofmM is

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a

16/9

b

3/5

c

5/3

d

25/9

answer is C.

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Detailed Solution

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T=2πMk when mass is increased by m then .....(i)  T'=2πM+mk 5T3=2πM+mk (ii) 

Dividing Eq. (i) by Eq. (ii), we get 

35=MM+m

925=MM+m

9M+9m=25M

16M=9m

mM=169

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