Question

A mass M when suspended by an ideal spring oscillates vertically with a time period of 1 second. If an additional mass of 3kg be also suspended with it, its time - period is doubled. Find the value of M.

Answer 1

$T=2\pi \sqrt{\frac{M}{k}}also,2T=2\pi \sqrt{\frac{M+3}{k}}$
$\Rightarrow 1=2\pi \sqrt{\frac{M}{k}}\to \left(1\right)\Rightarrow 2\pi \sqrt{\frac{M+3}{k}}\to \left(2\right)$
Dividing (2) by (1),
$2=\sqrt{\frac{M+3}{M}}\Rightarrow \frac{M+3}{M}\Rightarrow 4=\frac{M+3}{M}\Rightarrow 4M=M+3$
$\therefore M=1kg$