Q.28 State differential equation of linear S.H.M. hence obtain an expression for acceleration, velocity and displacement of a particle performing S.H.M.
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Answer:
Differential equation: In linear SHM the restoring force is \(F=-kx\), so \(m\dfrac{d^2x}{dt^2}=-kx\), i.e.
\(\dfrac{d^2x}{dt^2}+\omega^2x=0\), where \(\omega^2=\dfrac{k}{m}\).
Acceleration: From the equation, \(a=\dfrac{d^2x}{dt^2}=-\omega^2x\). Its magnitude is maximum (\(\omega^2A\)) at the extreme positions and zero at the mean position.
Velocity: Write \(a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=v\dfrac{dv}{dx}\). Then \(v\dfrac{dv}{dx}=-\omega^2x\Rightarrow v\,dv=-\omega^2x\,dx\).
Integrating: \(\dfrac{v^2}{2}=-\dfrac{\omega^2x^2}{2}+C\). At the extreme position \(x=\pm A\), \(v=0\), so \(C=\dfrac{\omega^2A^2}{2}\).
\(\therefore v^2=\omega^2(A^2-x^2)\Rightarrow v=\pm\omega\sqrt{A^2-x^2}\). Maximum speed \(\omega A\) at the mean position.
Displacement: \(v=\dfrac{dx}{dt}=\omega\sqrt{A^2-x^2}\Rightarrow\dfrac{dx}{\sqrt{A^2-x^2}}=\omega\,dt\).
Integrating: \(\sin^{-1}\left(\dfrac{x}{A}\right)=\omega t+\alpha\), where \(\alpha\) is a constant of integration (the initial phase).
\(\therefore x=A\sin(\omega t+\alpha)\).
Then \(v=\dfrac{dx}{dt}=A\omega\cos(\omega t+\alpha)\) and \(a=-A\omega^2\sin(\omega t+\alpha)=-\omega^2x\).
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