The velocity of a particle executing SHM varies with displacement $(x)$ as $4 v^{2}=50-x^{2}$. The time period of oscillations is $\frac{x}{7} s$. The value of $x$ is ___________. $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
Answer (integer)
88
Solution
<p>$4{v^2} = 50 - {x^2}$</p>
<p>or $v = {1 \over 2}\sqrt {50 - {x^2}}$</p>
<p>Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}}$</p>
<p>$\Rightarrow \omega = {1 \over 2}$</p>
<p>& $A = \sqrt {50}$</p>
<p>so ${{2\pi } \over T} = {1 \over 2}$</p>
<p>$\Rightarrow T = 4\pi \sec$</p>
<p>$= 4 \times {{22} \over 7}\sec$</p>
<p>$T = {{88} \over 7}\sec$</p>
<p>so $x = 88$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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