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A tray of mass $12 \mathrm{~kg}$ is supported by two identical springs as shown in figure. When the tray is pressed down slightly and then released, it executes SHM with a time period of $1.5 \mathrm{~s}$. The spring constant of each spring is
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Verified Answer
The correct answer is:
$105 \mathrm{Nm}^{-1}$
Mass of tray, $m=12 \mathrm{~kg}$
Time period, $T=1.5 \mathrm{~s}$
If $k$ be the spring constant of each spring, then
$k_{1}=k_{2}=k$
Since, springs are connected in parallel, hence
$\begin{aligned}
&k_{\text {net }}=k+k \\
&\text { Time period, } T=2 \pi \sqrt{\frac{m}{k_{\text {net }}}}=2 \pi \sqrt{\frac{m}{k_{1}+k_{2}}} \\
&\Rightarrow \quad 1.5=2 \pi \sqrt{\frac{12}{k+k}} \Rightarrow 1.5=2 \pi \sqrt{\frac{12}{2 k}} \\
&\Rightarrow \quad 1.5=2 \pi \sqrt{\frac{6}{k}}
\end{aligned}$
Squaring both side, we have
$\begin{aligned}
2.25 &=4 \pi^{2} \cdot \frac{6}{k} \\
\Rightarrow \quad k &=\frac{4 \pi^{2} \times 6}{2.25} \\
&=105.17 \mathrm{Nm}^{-1} \simeq 105 \mathrm{Nm}^{-1}
\end{aligned}$
Time period, $T=1.5 \mathrm{~s}$
If $k$ be the spring constant of each spring, then
$k_{1}=k_{2}=k$
Since, springs are connected in parallel, hence
$\begin{aligned}
&k_{\text {net }}=k+k \\
&\text { Time period, } T=2 \pi \sqrt{\frac{m}{k_{\text {net }}}}=2 \pi \sqrt{\frac{m}{k_{1}+k_{2}}} \\
&\Rightarrow \quad 1.5=2 \pi \sqrt{\frac{12}{k+k}} \Rightarrow 1.5=2 \pi \sqrt{\frac{12}{2 k}} \\
&\Rightarrow \quad 1.5=2 \pi \sqrt{\frac{6}{k}}
\end{aligned}$
Squaring both side, we have
$\begin{aligned}
2.25 &=4 \pi^{2} \cdot \frac{6}{k} \\
\Rightarrow \quad k &=\frac{4 \pi^{2} \times 6}{2.25} \\
&=105.17 \mathrm{Nm}^{-1} \simeq 105 \mathrm{Nm}^{-1}
\end{aligned}$
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