A light spring is suspended with mass ' m 1 ' at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of osci

Question

A light spring is suspended with mass ' m 1 ​ ' at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is T second. When a mass ' m 2 ​ ' is added to ' m 1 ​ ' and the system is made to oscillate the period is found to be 2 3 ​ T. The ratio ( m 2 ​ m 1 ​ ​ ) is

TestHub · Accepted Answer

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

For the initial case with mass $m_{1}$ :
$T = 2 π \frac{m _{1}}{k}$

When mass $m_{2}$ is added, the total mass becomes $m_{1} + m_{2}$ and the new time period is $\frac{3}{2} T$ :
$\frac{3}{2} T = 2 π \frac{m _{1} + m _{2}}{k}$

Dividing the second equation by the first equation:
$\frac{3}{2} = \frac{m _{1} + m _{2}}{m _{1}}$

Squaring both sides:
$\frac{9}{4} = \frac{m _{1} + m _{2}}{m _{1}}$

$\frac{9}{4} = 1 + \frac{m _{2}}{m _{1}}$

$\frac{m _{2}}{m _{1}} = \frac{9}{4} - 1 = \frac{5}{4}$

Taking the reciprocal to find the required ratio:
$\frac{m _{1}}{m _{2}} = \frac{4}{5}$

The ratio is $4 : 5$ .

Answer: $4 : 5$

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