A body of mass $4.9 \mathrm{~kg}$ hangs from a spring and oscillates with a period $0.5 \mathrm{~s}$. On the…
A body of mass $4.9 \mathrm{~kg}$ hangs from a spring and oscillates with a period $0.5 \mathrm{~s}$. On the removal of the body, the spring is shortened by (take, $g=10 \mathrm{~ms}^{-2}, \pi^2=10$ )
$6.3 \mathrm{~m}$
$0.63 \mathrm{~m}$
$6.25 \mathrm{~cm}$
$63 \mathrm{~cm}$
Solution
Given that, mass of body, $m=4.9 \mathrm{~kg}$
Time period of oscillation of spring,
$T=0.5 \mathrm{~s}$
Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$
and $\pi^2=10$
We know that, time period of spring
$T=2 \pi \sqrt{\frac{m}{k}}$
By squaring on both sides, $T^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}}$
$\Rightarrow \quad \frac{m}{k}=\frac{T^2}{4 \pi^2}$
$=\frac{(0.5)^2}{4 \times 10}=\frac{0.25}{40}=0.00625$
After removal of mass, length of spring decrease is equal to extension produced in spring.
By using equilibrium condition,
$F=m g$
$\Rightarrow \quad k x=m g$
$x=\frac{m}{k} g$
Substituting the values, we get
$\begin{aligned} x & =0.00625 \times 10 \\ & =0.0625 \mathrm{~m} \\ & =6.25 \mathrm{~cm}\end{aligned}$
Hence, the spring is shortened by $6.25 \mathrm{~cm}$ after removal of mass.