Question:medium

A sphere and a cube of equal masses on a horizontal frictionless floor, are confined between two vertical walls, as shown in the figure. The cube is attached to the wall by a massless spring. At the equilibrium position of the spring, the sphere just touches the cube. The cube is moved towards the left by a small amount $\ell$ from its equilibrium position, compressing the spring and is released at $t = 0$. The system keeps returning to its initial configuration as that of $t = 0$ with a time period $T$. If all the collisions are elastic, which of the following statements is correct?

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An increase in amplitude $\ell$ increases the maximum velocity of the cube ($v \propto \ell$).
Because the sphere travels at this higher speed, it spends less time covering the fixed distance to the wall, thereby reducing the overall period $T$.
Updated On: Jun 12, 2026
  • If $\ell$ increases, $T$ decreases.
  • If $\ell$ increases, $T$ does not change.
  • If $\ell$ increases, $T$ increases.
  • The sphere never moves.
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The Correct Option is A

Solution and Explanation


Step 1: Understanding the Concept:

The system undergoes a series of harmonic motions and elastic collisions. Because the masses are equal, they perfectly exchange velocities during elastic collisions.

Step 2: Detailed Explanation:

1. Sequence of Motion:
$\bullet$ Cube is released from \( x = -\ell \). It reaches \( x = 0 \) in time \( t_{1} = \frac{\pi}{2\omega} \).
$\bullet$ At \( x=0 \), it has max speed \( v = \omega \ell \). It hits the stationary sphere.
$\bullet$ Velocity exchange: Cube stops, sphere moves right with speed \( \omega \ell \).
$\bullet$ Sphere travels distance \( D \) to the wall and back. Time taken: \( t_{2} = \frac{2D}{\omega \ell} \).
$\bullet$ Sphere hits stationary cube at \( x=0 \). Velocity exchange: Sphere stops, cube moves left with \( \omega \ell \).
$\bullet$ Cube returns to \( x = -\ell \) in time \( t_{3} = \frac{\pi}{2\omega} \).
2. Total Time Period:
\[ T = \frac{\pi}{\omega} + \frac{2D}{\omega \ell} \]
3. Dependence on \( \ell \):
As \( \ell \) increases, the term \( \frac{2D}{\omega \ell} \) decreases while \( \frac{\pi}{\omega} \) remains constant. Therefore, the total period \( T \) decreases.

Step 3: Final Answer:

Increasing the initial compression \( \ell \) leads to a decrease in the time period T.
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