Question

When a particle of mass \(m\) is attached to a vertical spring of spring constant \(k\) and released, its motion is described by \(y(t)=y_0\sin^2\omega t\), where \(y\) is measured from the lower end of unstretched spring. Then \(\omega\) is :

• A. \(\sqrt{\dfrac{g}{y_0}}\)
• B. \(\dfrac{1}{2}\sqrt{\dfrac{g}{y_0}}\)
• C. \(\sqrt{\dfrac{2g}{y_0}}\)
• D. \(\sqrt{\dfrac{g}{2y_0}}\)

Hint:

Write \(\sin^2\omega t=\frac12(1-\cos2\omega t)\): the SHM has angular frequency \(2\omega\) about the mean position \(y_0/2\), which is the spring equilibrium extension \(mg/k\).

Answer 1

The Correct Option is D

\(y=y_0\sin^2\omega t=\dfrac{y_0}{2}-\dfrac{y_0}{2}\cos 2\omega t\).
So the particle oscillates about \(y=\dfrac{y_0}{2}\) with amplitude \(\dfrac{y_0}{2}\) and angular frequency \(\Omega=2\omega\).
The mean position is the equilibrium position of the loaded spring: \(\dfrac{y_0}{2}=\dfrac{mg}{k}\Rightarrow \dfrac{k}{m}=\dfrac{2g}{y_0}\).
For a spring-mass system \(\Omega=\sqrt{k/m}\), hence \(2\omega=\sqrt{\dfrac{2g}{y_0}}\).
\(\omega=\dfrac{1}{2}\sqrt{\dfrac{2g}{y_0}}=\sqrt{\dfrac{g}{2y_0}}\).
Answer: (4) \(\sqrt{\dfrac{g}{2y_0}}\)