A mass m is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period T . If the mass is increased by m 0 , the periodi

Question

A mass m is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period T . If the mass is increased by m 0 ​ , the periodic time becomes 4 5 T ​ . The ratio m m 0 ​ ​ is

TestHub · Accepted Answer

The time period of a spring-mass system is given by $T = 2 π \frac{m}{k}$ .

When the mass is increased by $m_{0}$ , the new time period is $T^{'} = 2 π \frac{m + m _{0}}{k}$ .

Given that $T^{'} = \frac{5 T}{4}$ , we have:

$2 π \frac{m + m _{0}}{k} = \frac{5}{4} (2 π \frac{m}{k})$

Squaring both sides, we get:

$\frac{m + m _{0}}{k} = \frac{25}{16} \frac{m}{k}$

$m + m_{0} = \frac{25}{16} m$

$m_{0} = \frac{25}{16} m - m = \frac{9}{16} m$

$\frac{m _{0}}{m} = \frac{9}{16}$

Answer: $\frac{9}{16}$

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