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When a body of mass $8 \mathrm{~kg}$ is attached to a spring balance, the reading of the balance is $20 \mathrm{~cm}$. Instead of $8 \mathrm{~kg}$, if another body of mass $M$ is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is $\frac{\pi}{5} \mathrm{~s}$, then the value of $\mathrm{M}$ is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$4 \mathrm{~kg}$
Mass of body, $\mathrm{m}=8 \mathrm{~kg}$
Period of oscillation, $\mathrm{T}=\frac{\pi}{5} \mathrm{~s}$
Reading of the balance, $x=20 \mathrm{~cm}=0.2 \mathrm{~m}$
For mass $8 \mathrm{~kg}$
$\mathrm{mg}=\mathrm{kx} \Rightarrow \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}$
$=\frac{8 \times 10}{0.2}=400 \mathrm{~N} / \mathrm{m}$
For mass $\mathrm{M}$
$\begin{aligned} & \omega^2=\frac{k}{M} \\ & M=\frac{k T^2}{4 \pi^2} \quad\left(\because \omega=\frac{2 \pi}{T}\right) \\ & =\frac{400 \times \pi^2}{4 \pi^2}=4 \mathrm{~kg}\end{aligned}$
Period of oscillation, $\mathrm{T}=\frac{\pi}{5} \mathrm{~s}$
Reading of the balance, $x=20 \mathrm{~cm}=0.2 \mathrm{~m}$
For mass $8 \mathrm{~kg}$
$\mathrm{mg}=\mathrm{kx} \Rightarrow \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}$
$=\frac{8 \times 10}{0.2}=400 \mathrm{~N} / \mathrm{m}$
For mass $\mathrm{M}$
$\begin{aligned} & \omega^2=\frac{k}{M} \\ & M=\frac{k T^2}{4 \pi^2} \quad\left(\because \omega=\frac{2 \pi}{T}\right) \\ & =\frac{400 \times \pi^2}{4 \pi^2}=4 \mathrm{~kg}\end{aligned}$
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