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 ( $\mu=$ coefficient of friction between the two blocks) (A) The time period of small oscillation of the two blocks is $\mathrm{T}=2 \pi \sqrt{\frac{(\mathrm{~m}+\mathrm{M})}{\mathrm{k}}}$ (B) The acceleration of the blocks is $\mathrm{a}=\frac{\mathrm{kx}}{\mathrm{M}+\mathrm{m}}$ ( $\mathrm{x}=$ displacement of the blocks from the mean position) (C) The magnitude of the frictional force on the upper block is $\frac{\mathrm{m} \mu|\mathrm{x}|}{\mathrm{M}+\mathrm{m}}$ (D) The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu(\mathrm{M}+\mathrm{m}) \mathrm{g}}{\mathrm{k}}$ (E) Maximum frictional force can be $\mu(M+m) g$. Choose the correct answer from the options given below :

Elasticity & SHM — JEE Main · 2025 · 03 Apr 2025 · Shift 1 Question with Solution

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{kx}{M + m}

(

x =

displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is

\frac{m μ ∣ x ∣}{M + m}

(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✓ Correct
(B)B, C, D Only
(C)C, D, E Only
(D)A, B, C Only

Answer: (A)

Solution

While the two blocks move together the oscillating mass is

M + m

, so the period is

T = 2 π \frac{M + m}{k}

and statement (A) holds. The spring force

k x

accelerates the whole system, so the common acceleration has magnitude

a = \frac{k x}{M + m}

and statement (B) holds. Friction on the upper block must supply exactly its share of this acceleration,

f = ma = \frac{mk x}{M + m}

; the expression in (C) carries a spurious factor

μ

and is wrong. The block does not slip as long as this required friction stays within the limit

μ m g

; setting

\frac{mk A}{M + m} = μ m g

gives the largest amplitude

A = \frac{μ ( M + m ) g}{k}

, so statement (D) holds. The greatest friction available between the blocks is

μ m g

, fixed by the normal force

m g

of the upper block, not

μ (M + m) g

, so (E) is wrong. The correct statements are A, B and D.

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Elasticity & SHM

Solution

More JEE Main questions on Elasticity & SHM