Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations1 Mark
Question
Show that for a particle executing S.H.M, velocity and displacement have a phase difference of $\frac{\pi}{2}.$
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Answer
Let the displacement equation of SHM$\text{x}=\text{a}\cos\omega\text{t}$
Velocity $\text{v}=\frac{\text{dx}}{\text{dt}}=\text{a}\omega(-\sin\omega\text{t})=-\text{a}\omega\sin\omega\text{t}$$\Rightarrow\text{v}=\text{a}\omega\cos\Big(\frac{\pi}{2}+\omega\text{t}\Big)$
Now, phase of displacement $\phi_1=\omega\text{t}$ Phase of velocity $\phi_2=\frac{\pi}{2}+\omega\text{t}$$\therefore$ Difference in phase of velocity to that of phase of displacement
$\triangle\phi=\phi_2-\phi_1=\Big(\frac{\pi}{2}+\omega\text{t}\Big)-(\omega\text{t})=\frac{\pi}{2}$
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