Physics · Oscillations · NEET
It DECREASES. For a spring, T = 2 pi square root of (m/k). k is fixed by the spring, so T depends only on m. Smaller m gives a smaller square root, so a shorter time period. The block oscillates faster. Many students wrongly copy the pendulum rule where mass does not matter at all - for a SPRING, mass DOES matter.
Sand leaks out mainly at the extreme positions, where the block is momentarily at rest (speed = 0, kinetic energy = 0). So no kinetic energy leaves with the sand - the mechanical energy E stays almost the same. Energy in SHM is E = (1/2) k A squared. Since k is fixed and E is unchanged, A stays fixed from this energy view; but writing E = (1/2) m omega squared A squared and remembering omega squared = k/m, the correct NEET conclusion (2025 paper) is that A INCREASES as m falls. Short version: mass falls, frequency rises, amplitude rises.
No. k is a property of the SPRING itself (its material, thickness and length), not of the load. Adding or removing mass never changes k. Only cutting the spring or combining springs changes k. So in a changing-mass problem, treat k as a constant and let only m vary.
No. A simple pendulum time period T = 2 pi square root of (L/g) has NO mass term - changing the bob mass does nothing to its period. A spring is the opposite: T = 2 pi square root of (m/k) depends fully on mass. Do not mix the two formulas.
Then m increases, so omega = square root of (k/m) DECREASES and T = 2 pi square root of (m/k) INCREASES. The block oscillates slower. This is the mirror image of the leaking case. Always start from omega = square root of (k/m) and see which way m moves.
In an oscillating spring-mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically, so the average angular frequency omega(t) and the average amplitude A(t) of the system change with time t. Which option correctly describes these changes?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
At any instant use omega = square root of (k/m) and T = 2 pi square root of (m/k), with k constant. As m changes, plug in the current mass. Smaller m gives larger omega and smaller T.
Frequency f = (1/2 pi) square root of (k/m). k does not change, so when m falls the fraction k/m grows, its square root grows, and f increases. The box oscillates faster over time.
No. k depends only on the spring (material, thickness, length). Changing the load never changes k. Only cutting or combining springs changes it.
A spring period depends on mass, T = 2 pi square root of (m/k), so mass matters. A pendulum period, T = 2 pi square root of (L/g), has no mass, so changing the bob mass does nothing.
Both the angular frequency omega(t) and the amplitude A(t) INCREASE with time. The correct option was D.
