- Home
- Standard 11
- Physics
A body of mass $m $ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released , it oscillates with a time period of $3\,s$ . When the mass $m$ is increased by $1\,kg$ , the time period of oscillations becomes $5\,s$ . The value of $m$ in $kg$ is
$\frac{16}{9}$
$\frac{9}{16}$
$\frac{3}{4}$
$\frac{4}{3}$
Solution
$Time\, period\, of\, spring – block \,system,$
$T=2 \pi \sqrt{\frac{m}{k}}$
For given spring, $T \propto \sqrt{m}$
$\frac{T_{1}}{T_{2}} =\sqrt{\frac{m_{1}}{m_{2}}}$
$\text { Here, } T_{1} =3 \mathrm{s}, m_{1}=m, T_{2}=5 \mathrm{s}, m_{2}=m+1, m=?$
$ \frac{3}{5}=\sqrt{\frac{m}{m+1}} \text { or } \frac{9}{25}=\frac{m}{m+1}$
$25 m=9 m+9 \Rightarrow 16 m=9$
$\therefore m=\frac{9}{16} \mathrm{kg}$
