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A pan with a set of weights is attached to a light spring. The period of vertical oscillation is 0.5 s . When some additional weights are put in pan, then the period of oscillations increases by 0.1 s . The extension caused by the additional weight is
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2.7 cm
Given, $T=0.5 \mathrm{~s}$,
We know that, $T=2 \pi \sqrt{\frac{m}{k}} \Rightarrow 0.5=2 \pi \sqrt{\frac{m}{k}}$
$\frac{1}{4 \pi}=\sqrt{\frac{m}{k}}$
On squaring both sides, we get
$\frac{m}{k}=\frac{1}{16 \pi^2}$...(i)
$\begin{aligned} & \Rightarrow \quad T+0.1=2 \pi \sqrt{\frac{m+m^{\prime}}{k}} \\ & \Rightarrow \quad 0.5+0.1=2 \pi \sqrt{\frac{m+m^{\prime}}{k}} \\ & \Rightarrow \quad\left(\frac{0.6}{2 \pi}\right)^2=\frac{m+m^{\prime}}{k}=\frac{m}{k}+\frac{m^{\prime}}{k} \\ & \end{aligned}$
$\Rightarrow \quad\left(\frac{0.6}{2 \pi}\right)^2=\frac{1}{16 \pi^2}+\frac{m^{\prime}}{k} \quad$ [using Eq. (i)]
$\Rightarrow \quad \frac{(0.6)^2}{4 \pi^2}-\frac{1}{16 \pi^2}=\frac{m^{\prime}}{k}$
$\Rightarrow \quad \frac{m^{\prime}}{k}=\frac{1.44-1}{16 \pi^2} \Rightarrow \frac{m^{\prime}}{k}=\frac{0.44}{16 \pi^2}$....(ii)
$\begin{aligned} \because & m^{\prime} g & =k x \\ \therefore & x & =\frac{m^{\prime}}{k} g\end{aligned}$
$=\frac{0.44}{16 \pi^2} \times 10 \quad$ [using Eq. (ii)]
$\begin{aligned} & =0.0279 \\ & =2.79 \times 10^{-2} \mathrm{~m} \\ & =2.7 \mathrm{~cm}\end{aligned}$
We know that, $T=2 \pi \sqrt{\frac{m}{k}} \Rightarrow 0.5=2 \pi \sqrt{\frac{m}{k}}$
$\frac{1}{4 \pi}=\sqrt{\frac{m}{k}}$
On squaring both sides, we get
$\frac{m}{k}=\frac{1}{16 \pi^2}$...(i)
$\begin{aligned} & \Rightarrow \quad T+0.1=2 \pi \sqrt{\frac{m+m^{\prime}}{k}} \\ & \Rightarrow \quad 0.5+0.1=2 \pi \sqrt{\frac{m+m^{\prime}}{k}} \\ & \Rightarrow \quad\left(\frac{0.6}{2 \pi}\right)^2=\frac{m+m^{\prime}}{k}=\frac{m}{k}+\frac{m^{\prime}}{k} \\ & \end{aligned}$
$\Rightarrow \quad\left(\frac{0.6}{2 \pi}\right)^2=\frac{1}{16 \pi^2}+\frac{m^{\prime}}{k} \quad$ [using Eq. (i)]
$\Rightarrow \quad \frac{(0.6)^2}{4 \pi^2}-\frac{1}{16 \pi^2}=\frac{m^{\prime}}{k}$
$\Rightarrow \quad \frac{m^{\prime}}{k}=\frac{1.44-1}{16 \pi^2} \Rightarrow \frac{m^{\prime}}{k}=\frac{0.44}{16 \pi^2}$....(ii)
$\begin{aligned} \because & m^{\prime} g & =k x \\ \therefore & x & =\frac{m^{\prime}}{k} g\end{aligned}$
$=\frac{0.44}{16 \pi^2} \times 10 \quad$ [using Eq. (ii)]
$\begin{aligned} & =0.0279 \\ & =2.79 \times 10^{-2} \mathrm{~m} \\ & =2.7 \mathrm{~cm}\end{aligned}$
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