Oscillations

JEE Main level, single correct, medium

133 of 194

Question 133 of 194

A particle executes simple harmonic motion with an amplitude of 5cm5cm . When the particle is at 4cm4cm from the mean position, the magnitude of its velocity in SISI units is equal to that of its acceleration. Then, its periodic time in seconds is:

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Answer: (A) 8π3\dfrac{8\pi }{3}

v=ω52−42=3ωv=\omega \sqrt{{5}^{2}-{4}^{2}}=3\omega

a=ω2×4a={\omega }^{2}\times 4

∣a∣=∣v∣\left|a\right|=\left|v\right|

4ω2=3ω4{\omega }^{2}=3\omega

ω=34=2πT\omega =\dfrac{3}{4}=\dfrac{2\pi }{T}

T=8π3secT=\dfrac{8\pi }{3}sec

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