TS EAMCET · Physics · Oscillations
A particle of mass 4 mg is executing simple harmonic motion along \(x\)-axis with an angular frequency of 40 rad \(\mathrm{s}^{-1}\). If the potential energy of the particle is \(V(x)=a+\) \(b x^2\), where \(V(x)\) is in joule and \(x\) is in metre, then the value of \(b\) is
- A \(800 \times 10^{-6} \mathrm{Jm}^{-2}\)
- B \(1600 \times 10^{-6} \mathrm{Jm}^{-2}\)
- C \(3200 \times 10^{-6} \mathrm{Jm}^{-2}\)
- D \(6400 \times 10^{-6} \mathrm{Jm}^{-2}\)
Answer & Solution
Correct Answer
(C) \(3200 \times 10^{-6} \mathrm{Jm}^{-2}\)
Step-by-step Solution
Detailed explanation
For particle executing SHM, \(\mathrm{m}=4 \times 10^{-3} \times 10^{-3} \mathrm{~kg}=4 \times 10^{-6} \mathrm{~kg}, \omega=40 \mathrm{rad} / \mathrm{s}\) potential energy, \(\mathrm{V}(\mathrm{x})=\mathrm{a}+\mathrm{bx} \mathrm{x}^2\) \(\therefore\) Restoring force,…
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