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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2022
  • Hard
  • Numerical

The potential energy of a particle of mass \(4\,kg\) in motion along the \(x\)-axis is given by \(U =4(1-\cos 4 x )\,J\). The time period of the particle for small oscillation \((\sin \theta \simeq \theta)\) is \(\left(\frac{\pi}{ K }\right)\,s\). The value of \(K\) is .......

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Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

\(U =4(1-\cos 4 x )\)

\(F =-\frac{ dU }{ dx }=-4(+\sin 4 x ) 4=-16 \sin (4 x )\)

For small \(\theta\)

\(\sin \theta \approx \theta\)

\(F=-64\,x\)

\(a=-64\,x / m=-16\,x\)

\(\omega^{2}=16\)

\(T =\frac{2 \pi}{\omega}=\frac{\pi}{2}\)

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