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 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$\frac{\mathrm{m}}{\mathrm{M}}$ is

1. A

3/5

2. B

25/9

3. C

16/9

4. D

5/3

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### Solution:

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

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

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

$⇒9\mathrm{M}+9\mathrm{m}=25\mathrm{M}$

$⇒16\mathrm{M}=9\mathrm{m}$

$⇒\frac{\mathrm{m}}{\mathrm{M}}=\frac{16}{9}$  Register to Get Free Mock Test and Study Material

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