Oscillations

JEE Main level, numerical answer, medium

6 of 194

Question 6 of 194

The velocity of a particle executing simple harmonic motion along xx-axis is described as v2=50−x2v^2 = 50 - x^2, where xx represents displacement. If the time period of motion is x7\dfrac{x}{7} s, the value of xx is _____.

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Answer: 44

The given equation is v2=50−x2v^2 = 50 - x^2

Differentiating with respect to xx:

2vdvdx=−2x2v \dfrac{dv}{dx} = -2x

vdvdx=−xv \dfrac{dv}{dx} = -x

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

a=−xa = -x

Comparing this with the standard equation of SHM a=−ω2xa = -\omega^2 x, we get:

ω2=1⇒ω=1\omega^2 = 1 \Rightarrow \omega = 1 rad/s

The time period of the motion is given by:

T=2πω=2πT = \dfrac{2\pi}{\omega} = 2\pi

Taking π=227\pi = \dfrac{22}{7}, we get:

T=2×227=447T = 2 \times \dfrac{22}{7} = \dfrac{44}{7} s

Given that the time period is x7\dfrac{x}{7} s, comparing the two gives:

x=44x = 44

Answer: 4444

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