Question

the period of oscillation of mass m suspended by an ideal spring is 2s. if an additional mass of 2kg be suspended the time period is increased by 1s. find the value of m.

Answer 1

The value of m is 1.6 kg.

Given:

Period of oscillation of m = 2 s

Additional mass = 2 kg

Period of oscillation of additional mass = 2 s + 1 s = 3 s

To find:

m = ?

Formula used:

Time period of oscillation:

$\bold{T=2\pi \sqrt{\frac{m}{k}}}$

Where,

k = Force constant

Solution:

Period of oscillation of m:

$\bold{T=2\pi \sqrt{\frac{m}{k}}}$

$\bold{2=2\pi \sqrt{\frac{m}{k}}\longrightarrow(1)}$

Period of oscillation of additional mass:

$\bold{3=2\pi \sqrt{\frac{(m+2)}{k}}\longrightarrow(2)}$

On dividing equation (1) by (2), we get,

$\bold{\frac{2}{3}=\frac{2\pi \sqrt{\frac{m}{k}}}{2\pi \sqrt{\frac{(m+2)}{k}}}}$

$\bold{\frac{2}{3}=\sqrt{\frac{\frac{m}{k}}{\frac{(m+2)}{k}}}}$

$\bold{\frac{4}{9}=\frac{m\times k}{(m+2)\times k}}$

$\bold{\frac{4}{9}=\frac{m}{m+2}}$

$\bold{4m+8=9m}$

$\bold{5m=8}$

$\bold{m=\frac{8}{5}}$

$\bold{\therefore m = 1.6 \ kg}$