MHT CET · Physics · Oscillations
A body of mass ' \(m\) ' performs linear S.H.M. given by equation \(\mathrm{x}=\mathrm{P} \sin \omega \mathrm{t}+\mathrm{Q} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)\). The total energy of the particle at any instant is
- A \(\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{PQ}\)
- B \(\frac{1}{2} \frac{\mathrm{m} \omega^2}{\mathrm{P}^2 \mathrm{Q}^2}\)
- C \(\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{P}^2+\mathrm{Q}^2\right)\)
- D \(\frac{1}{2} \mathrm{~m}^2 \mathrm{P}^2 \mathrm{Q}^2\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{P}^2+\mathrm{Q}^2\right)\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{x}=\mathrm{P} \sin \omega \mathrm{t}+\mathrm{Q} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)
\)
It can be considered as composition of two S.H.M. of amplitudes \(P\) and \(Q\) having phase difference \(\frac{\pi}{2}\).
\(\therefore\) Resultant amplitude \(\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}\)
Total energy \(E=\frac{1}{2} m \omega^2 R^2\)
\(
=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{P}^2+\mathrm{Q}^2\right)
\)
\mathrm{x}=\mathrm{P} \sin \omega \mathrm{t}+\mathrm{Q} \sin \left(\omega \mathrm{t}+\frac{\pi}{2}\right)
\)
It can be considered as composition of two S.H.M. of amplitudes \(P\) and \(Q\) having phase difference \(\frac{\pi}{2}\).
\(\therefore\) Resultant amplitude \(\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2}\)
Total energy \(E=\frac{1}{2} m \omega^2 R^2\)
\(
=\frac{1}{2} \mathrm{~m} \omega^2\left(\mathrm{P}^2+\mathrm{Q}^2\right)
\)
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