Oscillations

JEE Main level, single correct, medium

38 of 194

Question 38 of 194

An object of mass 0.5kg0.5kg is executing simple harmonic motion. It amplitude is 5cm5cm and time period (T) is 0.2s0.2s. What will be the potential energy of the object at an instant t=T4st=\dfrac{T}{4}s starting from mean position. Assume that the initial phase of the oscillation is zero.

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Answer: (A) 0.62J0.62J

T=2πmkT=2\pi \sqrt{\dfrac{m}{k}}
0.2=2π0.5k0.2=2\pi \sqrt{\dfrac{0.5}{k}}
k=50π2k=50{\pi }^{2}
≈500\approx 500
x=Asin⁡(ωt+ϕ)x=A\sin (\omega t+\phi )
=5cmsin(ωT4+0)=5cmsin\left(\dfrac{\omega T}{4}+0\right)
=5cmsin(π2)=5cmsin\left(\dfrac{\pi }{2}\right)
=5cm=5cm
PE=12kx2PE=\dfrac{1}{2}{kx}^{2}
=12(500)(5100)2=\dfrac{1}{2}(500){\left(\dfrac{5}{100}\right)}^{2}
=0.6255=0.6255

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