JEE Advanced · Physics · 15. Oscillations
Paragraph :
A uniform thin cylindrical disk of mass \(M\) and radius \(R\) is attached to two identical massless springs of spring constant \(k\) which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance \(d\) from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is \(L\). The disk is initially at its equilibrium position with its centre of mass \((C M)\) at a distance Lfrom the wall. The disk rolls without slipping with velocity \(\mathbf{v}_0=v_0 \hat{\mathbf{i}}\) The coefficient of friction is \(\mu\).
Question :
The net external force acting on the disk when its centre of mass is at displacement \(x\) with respect to its equilibrium position is
- A \(-k x\)
- B \(-2 k x\)
- C \(-\frac{2 k x}{3}\)
- D \(-\frac{4 k x}{3}\)
Answer & Solution
Correct Answer
(D) \(-\frac{4 k x}{3}\)
Step-by-step Solution
Detailed explanation
\(\therefore \frac{2 k x-f}{M}=R\left[\frac{f \cdot R}{\frac{1}{2} M R^2}\right]\)
Solving this equation, we get \(f=\frac{2 k x}{3}\)
\(
\therefore\left|F_{\text {net }}\right|=2 k x-f=2 k x-\frac{2 k x}{3}=\frac{4 k x}{3}
\)
This is opposite to displacement.
\(
\therefore F_{\text {net }}=-\frac{4 k x}{3}
\)
\(\therefore\) correct option is (d).
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