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 .......
Pick an option to check your answer.
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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