Column I shows spring block system with a constant force permanently acting on block. Match entries of column-I with column-II. Column I Column II (A) spring is initially relaxed when force is applied

Question

Column I shows spring block system with a constant force permanently acting on block. Match entries of column-I with column-II.

Column I	Column II
(A) spring is initially relaxed when force is applied	(P) Time period of oscillation $T = 2 π \frac{m}{k}$
(B) spring is initially relaxed when force is applied	(Q) Amplitude of oscillation is $A = \frac{2 m g}{k}$
(C) Before force is applied block is in equilibrium position	(R) Maximum velocity attained by block is $2 g [\frac{m}{k}]$
(D) Before force is applied block is in equilibrium position.	(S) Maximum magnitude of acceleration of block is 2g.
	(T) Velocity of block when spring is in natural length is zero. (If spring acquire natural length).

TestHub · Accepted Answer

(A) For equilibrium: 2 m g = k x x = k 2 m g ​ A = k 2 m g ​ v ma x ​ = A ω = ( k 2 m g ​ ) m k ​ ​ = 2 g k m ​ ​ a ma x ​ = ω 2 A = ( m k ​ ) × ( k 2 m g ​ ) = 2 g --- (B) Same as (A) --- (C) Initially: m g = k x 0 ​ 3 m g = k x A = k 2 m g ​ Spring will not acquire natural length. --- (D) m g = k x 0 ​ Spring is compressed by k m g ​ in equilibrium. For the block: k x = m g (upward) Weight = m g (downward) Applied force F = 2 m g (upward) So, x = k m g ​ Amplitude: A = k 2 m g ​ Velocity at natural length: v = ω A 2 − x 2 ​ where x = k m g ​

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Advanced 2027 (Class 12)