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As shown in the figure, a block of weight $20 \mathrm{~N}$ is connected to the top of a smooth inclined plane by massless spring of constant $8 \pi^2 \mathrm{Nm}^{-1}$. If the block is pulled slightly from its mean position and released, the period of oscillations 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:
$1 \mathrm{~s}$
Weight of block, $\mathrm{F}=20 \mathrm{~N}$
Spring constant, $\mathrm{K}=8 \pi^2 \mathrm{Nm}^{-1}$
$\begin{aligned}
& \mathrm{mg}=20 \Rightarrow \mathrm{m}=\frac{20}{10}=2 \mathrm{~kg} \\
& \omega=\frac{\mathrm{K}}{\mathrm{m}}=\frac{8 \pi^2}{2}=4 \pi^2 \\
& \frac{4 \pi^2}{\mathrm{~T}^2}=4 \pi^2 ; \mathrm{T}=1 \mathrm{~s}
\end{aligned}$
Period of oscillations, $\mathrm{T}=1 \mathrm{~s}$
Spring constant, $\mathrm{K}=8 \pi^2 \mathrm{Nm}^{-1}$
$\begin{aligned}
& \mathrm{mg}=20 \Rightarrow \mathrm{m}=\frac{20}{10}=2 \mathrm{~kg} \\
& \omega=\frac{\mathrm{K}}{\mathrm{m}}=\frac{8 \pi^2}{2}=4 \pi^2 \\
& \frac{4 \pi^2}{\mathrm{~T}^2}=4 \pi^2 ; \mathrm{T}=1 \mathrm{~s}
\end{aligned}$
Period of oscillations, $\mathrm{T}=1 \mathrm{~s}$
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