MHT CET · Physics · Oscillations
For a particle performing S.H.M., when displacement is 'x', the potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between \(\mathrm{x}, \mathrm{E}\) and \(\mathrm{F}\) is
- A \(\frac{E}{F}+x=0\)
- B \(\frac{2 E}{F}+x=0\)
- C \(\frac{E}{F}-x=0\)
- D \(\frac{2 E}{F}-x=0\)
Answer & Solution
Correct Answer
(B) \(\frac{2 E}{F}+x=0\)
Step-by-step Solution
Detailed explanation
Displacement \(=x, \quad\) P.E. \(=E, \quad\) Force \(=F\)
\(F=-k x\)
\(E=\frac{1}{2} m \omega^{2} x^{2}=\frac{1}{2} k x^{2}\)
\(2 E=(-) \frac{F}{x} \times x^{2}=(-) F x\)
\(\frac{2 E}{F}+x=0\)
\(F=-k x\)
\(E=\frac{1}{2} m \omega^{2} x^{2}=\frac{1}{2} k x^{2}\)
\(2 E=(-) \frac{F}{x} \times x^{2}=(-) F x\)
\(\frac{2 E}{F}+x=0\)
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