The velocity of a particle executing simple… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Waves & Oscillations

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The velocity of a particle executing simple harmonic motion along $x$ -axis is described as $v^2 = 50 - x^2$ , where $x$ represents displacement. If the time period of motion is $\dfrac{x}{7}$ s, the value of $x$ is _____.

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

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Solution

The given equation is $v^2 = 50 - x^2$

Differentiating with respect to $x$ :

$2v \dfrac{dv}{dx} = -2x$

$v \dfrac{dv}{dx} = -x$

Since acceleration $a = v \dfrac{dv}{dx}$ , we get:

$a = -x$

Comparing this with the standard equation of SHM $a = -\omega^2 x$ , we get:

$\omega^2 = 1 \Rightarrow \omega = 1$ rad/s

The time period of the motion is given by:

$T = \dfrac{2\pi}{\omega} = 2\pi$

Taking $\pi = \dfrac{22}{7}$ , we get:

$T = 2 \times \dfrac{22}{7} = \dfrac{44}{7}$ s

Given that the time period is $\dfrac{x}{7}$ s, comparing the two gives:

$x = 44$

Answer: $44$

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