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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2021
  • Medium
  • MCQ

An object of mass \(0.5\, {kg}\) is executing simple harmonic motion. Its amplitude is \(5\, {cm}\) and time period (T) is \(0.2\, {s} .\) What will be the potential energy of the object at an instant \(t=\frac{T}{4}\) s starting from mean position. Assume that the initial phase of the oscillation is zero. (In \({J}\))

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Answer & Solution

Correct Answer

(B) \(0.62\)

Step-by-step Solution

Detailed explanation

by using formula of time period

\(T=2 \pi \sqrt{\frac{m}{k}}\)

\(0.2=2 \pi \sqrt{\frac{0.5}{k}}\)

\(k=50 \pi^{2}\)

\(\approx 500\)

\(x=A\, \sin (\omega t+\phi)\)

\(=5\, c m\, \sin \left(\frac{\omega t}{4}+0\right)\)

\(=5\, {cm} \sin \left(\frac{\pi}{2}\right)\)

\(=5 \,{cm}\)

\(P E=\frac{1}{2} k x^{2}\)

\(=\frac{1}{2}(500)\left(\frac{5}{100}\right)^{2}\)

\(=0.6255\)

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