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A mass of $5 \mathrm{~kg}$ is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length $4 \mathrm{~m}$ has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?
Option: 1 $\begin{aligned} &4 \mathrm{~m} / \mathrm{s}^{2} \\ \end{aligned}$
Option: 2 $9.8 \mathrm{~m} / \mathrm{s}^{2} \\$
Option: 3 $5 \mathrm{~m} / \mathrm{s}^{2} \\$
Option: 4 $10 \mathrm{~m} / \mathrm{s}^{2}$

Answers (1)

best_answer

$U_{max}= \frac{1}{2}m\omega ^{2}A^{2}= 10J$
$\frac{1}{2}\times 5\times \omega ^{2}\times \left ( 2 \right )^{2}= 10$
$\omega ^{2}= 1$
$\omega = 1$
$T= \frac{2\pi }{\omega }= 2\pi s$
For simple pendulum,
$Time \, period= 2\pi \sqrt{\frac{l}{g}}= 2\pi$
$l= g$
$g= 4\frac{m}{s^{2}}$
Accelaration due to gravity on that planet is $4\frac{m}{s^{2}}$
The correct option is (1)

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Deependra Verma

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