If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
Solution
<p>The formula for the period of a simple pendulum is given by:</p>
<p>$T = 2\pi\sqrt{\frac{l}{g}}$</p>
<p>where:</p>
<ul>
<li>T is the period of the pendulum,</li>
<li>l is the length of the pendulum, and</li>
<li>g is the acceleration due to gravity.</li>
</ul>
<p>We need to find g. Rearranging the formula for g, we get:</p>
<p>$g = \frac{4\pi^2l}{T^2}$</p>
<p>Given:</p>
<ul>
<li>l = 2 m,</li>
<li>T = 2 s,</li>
</ul>
<p>Substituting these values into the equation, we get:</p>
<p>$g = \frac{4\pi^2 \times 2}{(2)^2} = 2\pi^2 \, ms^{-2}$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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