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NCERT Exemplar · Q13

Q.Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is (Note: more than one of the given options may be correct.)

(a) simple harmonic motion.
(b) non-periodic motion.
(c) periodic motion.
(d) periodic but not S.H.M.
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For small displacements from the bottom of a smooth curved bowl, the restoring force is proportional to displacement, producing simple harmonic motion. For larger displacements the motion remains periodic but the proportionality breaks down, making it non-SHM.

Why this matters: the geometry of restoring forces

When a ball bearing sits at the lowest point of a smooth bowl, it is in equilibrium. Displace it slightly to one side and gravity pulls it back. The key question is: how does the restoring force depend on displacement?

For simple harmonic motion we need the restoring force to be strictly proportional to displacement from equilibrium: F=−kxF = -kx. This happens when the bowl's curvature near the bottom is approximately circular (or parabolic in the vertical cross-section). For small angles, sin⁡θ≈θ\sin\theta \approx \theta, and the component of gravity along the surface becomes proportional to the arc displacement.

For larger displacements, the small-angle approximation fails. The restoring force is still directed toward equilibrium and the ball still oscillates back and forth, but the force is no longer linear in displacement. The motion remains periodic—it repeats—but the period depends on amplitude, which violates the defining property of SHM.

Step-by-step analysis

  1. Small displacement regime Release the ball bearing from a point "slightly above" the lower point. If the bowl is smooth (frictionless) and has a regular curved profile, the tangential component of gravitational force is

Ftangent=−mgsin⁡θF_{\text{tangent}} = -mg\sin\theta

where θ\theta is the angle the surface makes with the horizontal at the ball's position.

  1. Small-angle approximation For small θ\theta, sin⁡θ≈θ≈sR\sin\theta \approx \theta \approx \frac{s}{R}, where ss is the arc displacement from the bottom and RR is the radius of curvature. Then

F≈−mgsR=−(mgR)sF \approx -mg\frac{s}{R} = -\left(\frac{mg}{R}\right)s

This is a linear restoring force, the hallmark of SHM. The angular frequency is ω=g/R\omega = \sqrt{g/R} and the motion is sinusoidal in time.

  1. Checking the options for small displacement

    • (A) Simple harmonic motion: Yes, the force is proportional to displacement.
    • (C) Periodic motion: Yes, SHM is a special case of periodic motion.
    • (B) Non-periodic: No, the motion repeats.
    • (D) Periodic but not SHM: No, it is SHM in this regime.
  2. Larger displacement regime …

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