The upper end of the spring is fixed and a mass ' $\mathrm{m}$ ' is attached to its lower end. When mass is…
The upper end of the spring is fixed and a mass ' $\mathrm{m}$ ' is attached to its lower end. When mass is slightly pulled down and released, it oscillates with time period 3 second. If mass ' $\mathrm{m}$ ' is increased by $1 \mathrm{~kg}$, the time period becomes 5 second. The value of ' $\mathrm{m}$ ' is (mass of spring is negligible)
$\frac{3}{8} \mathrm{~kg}$
$\frac{5}{9} \mathrm{~kg}$
$\frac{8}{13} \mathrm{~kg}$
$\frac{9}{16} \mathrm{~kg}$
Solution
The formula for the time period of a spring mass system is $T=2 \pi \sqrt{\frac{m}{k}}$
For mass $m+1, T^{\prime}=2 \pi \sqrt{\frac{m+1}{k}}$
Taking the ratio,
$\begin{aligned}
& \frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\frac{2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}}{2 \pi \sqrt{\frac{\mathrm{m}+1}{\mathrm{k}}}} \\
& \frac{\mathrm{T}}{\mathrm{T}^{\prime}}=\sqrt{\frac{\mathrm{m}}{\mathrm{m}+1}} \\
& \frac{3}{5}=\sqrt{\frac{m}{m+1}} \\
& \frac{9}{25}=\frac{m}{m+1} \\
& \therefore \quad 9 m+9=25 \mathrm{~m} \\
& \therefore 16 \mathrm{~m}=9 \\
& \therefore \quad \mathrm{m}=\frac{9}{16} \mathrm{~kg} \\
&
\end{aligned}$