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 \({J}\))
Pick an option to check your answer.
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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