Oscillations

JEE Main level, numerical answer, medium

114 of 194

Question 114 of 194

The velocity of a particle executing SHM varies with displacement (xx) as 4v2=50–x24{v}^{2}=50–{x}^{2}. The time period of oscillations is x7s\dfrac{x}{7}s. The value of xx is ______.
[Take π=227\pi =\dfrac{22}{7}]

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Answer: 88

Given:
4v2=50−x24{v}^{2}=50-{x}^{2}
⇒v=1250−x2\Rightarrow v=\dfrac{1}{2}\sqrt{50-{x}^{2}}
Comparing with standard equation of SHM, v=ωA2−x2v=\omega \sqrt{{A}^{2}-{x}^{2}}, we get
ω=12\omega =\dfrac{1}{2}
Now,
T=2πω=4π=887T=\dfrac{2\pi }{\omega }=4\pi =\dfrac{88}{7}
Hence, required x=88x=88.

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