Question

A mass $M$ attached to a spring oscillates with a period of $2$ seconds. If the mass is increased by $2kg$, the period increases by $1$ second. Find the initial mass, assuming that Hooke's law is obeyed.

Answer 1

Given $m_1=M$
$T_1=2s$
$m_2=M+2kg$
$T_2=2s+1s=3s$
Period $T=2\pi \sqrt{\frac{m}{K}}$ (in the usual notation)

$\therefore$ $T_1=2\pi \sqrt{\frac{m}{K}}$ and $T_2=2\pi \sqrt{\frac{M+2}{K}}$

$\therefore$ $\frac{T_2}{T_1}=\sqrt{\frac{M+2}{M}}$

$\frac{3}{2}=\sqrt{\frac{M+2}{M}}$

$\frac{9}{4}=\frac{M+2}{M}$

$9M=4M+8$

$9M-4M=8$

$5M=8$

$M=\frac{8}{5}=1.6kg$