JEE Main 2025 Apr 3 Shift 1, Physics Q6: Simple Harmonic Motion And Oscillation

JEE Main2025Apr 3, Shift 1Physics

Two blocks of masses $m$ and $M$ , ( $M > m$ ), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released then ( $μ$ = coefficient of friction between the two blocks)

(A) The time period of small oscillation of the two blocks is $T = 2 π \frac{m + M}{k}$

(B) The acceleration of the blocks is $a = \frac{k x}{M + m}$ ( $x$ = displacement of the blocks from the mean position)

(D) The maximum amplitude of the upper block, if it does not slip, is $\frac{μ ( M + m ) g}{k}$

(E) Maximum frictional force can be $μ (M + m) g$ .

Choose the correct answer from the options given below :

A
A, B, D Only
B
B, C, D Only
C
C, D, E Only
D
A, B, C Only

Solution

Both blocks move together (no slip), so treat the system as a single oscillator of total mass $M + m$ attached to a spring of constant $k$ .

(A) $T = 2 π \frac{M + m}{k}$ . ✓

(B) Restoring force is $- k x$ on the combined mass, so $a = \frac{k x}{M + m}$ (magnitude). ✓

(C) Friction is the only horizontal force on the upper block, providing its acceleration: $f = ma = \frac{mk x}{M + m}$ . This does not contain $μ$ . The expression in (C) contains an extra $μ$ factor, so (C) is incorrect.

(D) The upper block can be carried along only as long as required friction $\leq μ m g$ . Setting $\frac{mk A _{m a x}}{M + m} = μ m g$ gives $A_{m a x} = \frac{μ ( M + m ) g}{k}$ . ✓

(E) The normal force on the upper block is $m g$ , so maximum static friction is $μ m g$ , not $μ (M + m) g$ . (E) is incorrect.

$∴$ Correct statements: A, B, D.

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JEE Main 2025 Apr 3 Shift 1, Physics Q6: Simple Harmonic Motion And Oscillation

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