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

A
3/5
B
25/9
C
16/9
D
5/3

Solution:

$T = 2 π \sqrt{\frac{M}{k}} when mass is increased by m then . . . . . (i) T' = 2 π \sqrt{\frac{M + m}{k}} \Rightarrow \frac{5 T}{3} = 2 π \sqrt{\frac{M + m}{k}} \dots \dots (ii)$

Dividing Eq. (i) by Eq. (ii), we get

$\frac{3}{5} = \sqrt{\frac{M}{M + m}}$

$\frac{9}{25} = \frac{M}{M + m}$

$\Rightarrow 9 M + 9 m = 25 M$

$\Rightarrow 16 M = 9 m$

$\Rightarrow \frac{m}{M} = \frac{16}{9}$

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 of $\frac{m}{M}$ is

Solution:

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