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 $T$.
Step 1: Understanding the Question:
We are given a physical system consisting of a spring-mass cube of mass $m$ and a sphere of equal mass $m$ on a frictionless floor between two walls.
The cube is initially compressed by a distance $\ell$ and released.
Since all collisions (cube-sphere and sphere-wall) are elastic, we need to trace the motion of both objects over one complete cycle and determine how the total time period $T$ depends on the initial displacement $\ell$. Step 2: Key Formula or Approach:
For simple harmonic motion (SHM) of the cube-spring system, the angular frequency is $\omega = \sqrt{\frac{k}{m}}$, and the displacement is described by $x(t) = -\ell \cos(\omega t)$.
For an elastic collision between two equal masses where one is initially moving and the other is at rest, the two bodies completely exchange their velocities.
The time period $T$ of the system is the sum of the time spent by the cube in simple harmonic motion and the time spent by the sphere traveling back and forth. Step 3: Detailed Explanation:
• The cube is released from $x = -\ell$ at $t = 0$ and reaches the equilibrium position $x = 0$ at $t_1 = \frac{\pi}{2\omega}$.
• The velocity of the cube at $x = 0$ is the maximum velocity of SHM:
\[ v_{max} = \omega \ell \]
• At $x = 0$, the cube collides elastically with the stationary sphere of equal mass $m$.
• Due to the equal mass elastic collision, the cube comes to rest, and the sphere moves to the right with velocity $v_{sphere} = \omega \ell$.
• Let $D$ be the distance from the equilibrium position to the right wall.
• The sphere travels to the wall, collides elastically with it (reversing its velocity), and returns to $x = 0$.
• The time taken by the sphere for this round trip is:
\[ t_2 = \frac{2D}{v_{sphere}} = \frac{2D}{\omega \ell} \]
• When the sphere returns to $x = 0$, it collides elastically with the stationary cube, transferring all its momentum back to the cube.
• The sphere comes to rest, and the cube moves to the left with velocity $\omega \ell$.
• The cube goes to the extreme left position $x = -\ell$ and comes back to $x=0$, which takes a half-period of SHM:
\[ t_3 = \frac{\pi}{\omega} \]
• The total time period $T$ for one complete cycle is:
\[ T = t_{cube} + t_{sphere} = \frac{\pi}{\omega} + \frac{2D}{\omega \ell} \]
• Looking at the expression for $T$, as $\ell$ increases, the term $\frac{2D}{\omega \ell}$ decreases while $\frac{\pi}{\omega}$ remains constant.
• Therefore, as $\ell$ increases, the total time period $T$ decreases. Step 4: Final Answer:
Thus, if $\ell$ increases, $T$ decreases.
