Question

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 \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)

Choose the correct answer from the options given below:

• A. A, B and D
• B. B, C and D
• C. A, E and D
• D. C, D, and E

Hint:

Analyze the forces acting on the blocks and apply Newton's second law to determine the acceleration and frictional force. Consider the conditions for maximum amplitude and maximum frictional force.

Answer 1

The Correct Option is A

This problem analyzes a two-block system (masses m and M) on a frictionless table, with m on top of M. Block M is attached to a spring (constant k). A coefficient of static friction \( \mu \) exists between m and M. The system undergoes simple harmonic motion (SHM). We must identify all correct statements among five given options.

Concepts Employed:

The problem integrates principles of Simple Harmonic Motion (SHM), Newton's Second Law, and static friction.

1. Simple Harmonic Motion (SHM): SHM occurs when the net restoring force is directly proportional to displacement from equilibrium and directed towards it: \( F = -K_{eff} x \). The time period is:

\[ T = 2\pi\sqrt{\frac{M_{total}}{K_{eff}}} \]

2. Newton's Second Law: The net force on an object equals its mass times acceleration: \( F_{net} = ma \). This applies to the combined system and individual blocks.

3. Static Friction: Static friction prevents relative motion. Its magnitude opposes impending motion and can reach a maximum:

\[ f_{s,max} = \mu N \]

where \( N \) is the normal force. For block m, \( N = mg \).

Step-by-Step Analysis:

Each statement is evaluated assuming the blocks oscillate together without slipping, unless specified otherwise.

Statement (A): Time Period

When blocks move together, their combined mass is \( (m+M) \). The spring provides the restoring force \( -kx \).

Applying Newton's Second Law to the combined system:

\[ F_{net} = (m+M)a = -kx \]

The equation of motion \( a = -\frac{k}{m+M}x \) is of the form \( a = -\omega^2 x \). Thus, the angular frequency is:

\[ \omega = \sqrt{\frac{k}{m+M}} \]

The time period is \( T = \frac{2\pi}{\omega} \):

\[ T = 2\pi\sqrt{\frac{m+M}{k}} \]

This confirms Statement (A). Statement (A) is correct.

Statement (B): Acceleration

From the equation of motion derived above:

\[ a = -\frac{kx}{m+M} \]

The magnitude of acceleration is:

\[ a = \frac{k|x|}{M+m} \]

This confirms Statement (B). Statement (B) is correct.

Statement (C): Frictional Force

The upper block (m) accelerates with the lower block. The horizontal force on m is static friction \( f_s \).

Applying Newton's Second Law to block m:

\[ f_s = ma \]

Substituting the acceleration from Step 2:

\[ f_s = m \left( \frac{k|x|}{M+m} \right) = \frac{mk|x|}{M+m} \]

Statement (C) incorrectly states the frictional force as \( \frac{m\mu|x|}{M+m} \). The force depends on acceleration, not directly on \( \mu \) unless it reaches maximum. Statement (C) is incorrect.

Statement (D): Maximum Amplitude Without Slipping

Slipping occurs when the required acceleration force exceeds maximum static friction.

Required force \( f_s = ma = m \omega^2 |x| \). Maximum at \( |x| = A_{max} \):

\[ f_{s, required} = m \omega^2 A_{max} \]

Maximum static friction is \( f_{s,max} = \mu N = \mu mg \).

For no slipping, \( f_{s, required} \le f_{s,max} \). At the limit:

\[ m \omega^2 A_{max} = \mu mg \]\[ A_{max} = \frac{\mu g}{\omega^2} \]

Substituting \( \omega^2 = \frac{k}{m+M} \):

\[ A_{max} = \frac{\mu g}{k/(m+M)} = \frac{\mu(m+M)g}{k} \]

This matches Statement (D). Statement (D) is correct.

Statement (E): Maximum Frictional Force

The maximum static friction between m and M depends on the normal force between them.

Normal force \( N = mg \).

Maximum static friction is:

\[ f_{s,max} = \mu N = \mu mg \]

Statement (E) incorrectly states \( \mu(M+m)g \). This expression relates to friction with the table, not between the blocks. Statement (E) is incorrect.

Conclusion:

The correct statements are (A), (B), and (D).

Comparing with the options:

(1) A, B, D Only

(2) B, C, D Only

(3) C, D, E Only

(4) A, B, C Only

Option (1) matches the correct statements.

The correct answer is (1) A, B, D Only.