WBJEE · Physics · Oscillations
The potential energy of a particle of mass \(0.02 \mathrm{~kg}\) moving along \(x\)-axis is given by \(V=A x(x-4) J\) where \(x\) is in metres and \(\mathrm{A}\) is a constant. Which of the following is/are correct statement(s)?
- A The particle is acted upon by a constant force
- B The particle executes simple harmonic motion
- C The speed of the particle is maximum at \(x=2 m\)
- D The period of oscillation of the particle is \(\frac{\pi}{5} \mathrm{sec}\)
Answer & Solution
Correct Answer
(C) The speed of the particle is maximum at \(x=2 m\)
Step-by-step Solution
Detailed explanation
\(V=A x(x-4) J m=0.02=A x^{2}-4 A x\) \[ \begin{array}{l} \mathrm{F}=-\frac{\mathrm{d} \mathrm{V}}{\mathrm{dx}}=-2 \mathrm{Ax}+4 \mathrm{~A} \text { (not constant) } \\ \mathrm{F}=-2 \mathrm{~A}(\mathrm{x}-2) \end{array} \] Hence position execute S.H.M mean position is when…
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