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

The problem describes a system of two blocks, with masses m and M, where m is placed on top of M. The block M is attached to a spring with constant k and rests on a frictionless table. There is a coefficient of static friction μ between the two blocks. The system undergoes simple harmonic motion (SHM). We need to evaluate five statements about the motion and choose the option that lists all the correct statements.

Concept Used:

This problem involves the principles of Simple Harmonic Motion (SHM), Newton's Second Law, and the concept of static friction.

1. Simple Harmonic Motion (SHM): For a system to execute SHM, the net restoring force must be directly proportional to the displacement from the mean position and directed towards it. \( F = -K_{eff} x \). The time period of such a system is given by:

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

2. Newton's Second Law: The net force on an object is equal to the product of its mass and acceleration \( (F_{net} = ma) \). This will be applied to the two-block system as a whole and to the individual upper block.

3. Static Friction: Static friction is a self-adjusting force that prevents relative motion between surfaces. It acts in the direction opposite to the impending motion. Its magnitude is just enough to provide the necessary acceleration, up to a maximum limit given by:

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

where \( \mu \) is the coefficient of static friction and \( N \) is the normal force between the surfaces. For the upper block, \( N = mg \).

Step-by-Step Solution:

We will analyze each statement individually, assuming the two blocks oscillate together without slipping, unless stated otherwise.

Step 1: Analysis of Statement (A)

Statement (A) gives the time period of oscillation. When the two blocks move together, they can be treated as a single system of total mass \( (m+M) \). The restoring force is provided by the spring and is equal to \( -kx \), where x is the displacement from the mean position.

Applying Newton's Second Law to the combined system:

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

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

\[ \omega^2 = \frac{k}{m+M} \implies \omega = \sqrt{\frac{k}{m+M}} \]

The time period T is related to the angular frequency by \( T = \frac{2\pi}{\omega} \).

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

This matches Statement (A). Therefore, Statement (A) is correct.

Step 2: Analysis of Statement (B)

Statement (B) gives the acceleration of the blocks. From the equation of motion derived in Step 1:

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

The magnitude of the acceleration is:

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

This matches Statement (B). Therefore, Statement (B) is correct.

Step 3: Analysis of Statement (C)

Statement (C) gives the magnitude of the frictional force on the upper block. The upper block (mass m) accelerates along with the lower block (mass M). The only horizontal force acting on the upper block is the static friction \( f_s \) from the lower block. Applying Newton's Second Law to the upper block:

\[ f_s = ma \]

Substituting the expression for acceleration 'a' from Step 2:

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

Statement (C) claims the frictional force is \( \frac{m\mu|x|}{M+m} \), which is incorrect. The frictional force depends on the acceleration (and thus on k), not directly on the coefficient of friction \( \mu \), unless it is at its maximum value. Therefore, Statement (C) is incorrect.

Step 4: Analysis of Statement (D)

Statement (D) asks for the maximum amplitude for which the upper block does not slip. Slipping will occur when the force required to accelerate the upper block exceeds the maximum possible static frictional force.

The required force is \( f_s = ma = m \omega^2 |x| \). This force is maximum at the extreme position, where \( |x| = A_{max} \).

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

The maximum available static frictional force is \( f_{s,max} = \mu N = \mu mg \).

For no slipping, we must have \( f_{s, required} \le f_{s,max} \). The limiting condition is:

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

Substituting \( \omega^2 = \frac{k}{m+M} \) from Step 1:

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

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

Step 5: Analysis of Statement (E)

Statement (E) claims the maximum frictional force is \( \mu(M+m)g \). The frictional force acts between the upper block (m) and the lower block (M). The maximum value of this static friction depends on the normal force between these two surfaces.

The normal force on the upper block is equal to its weight, \( N = mg \).

Therefore, the maximum possible static frictional force is:

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

The expression \( \mu(M+m)g \) represents the normal force from the table on the lower block M multiplied by \( \mu \), which is not relevant to the friction between m and M. Therefore, Statement (E) is incorrect.

Final Computation & Result:

Based on our analysis, the correct statements are (A), (B), and (D).

We now check the given options:

(1) A, B, D Only

(2) B, C, D Only

(3) C, D, E Only

(4) A, B, C Only

The combination of correct statements matches option (1).

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

Answer 2

(A) As both blocks moving together so time period = \( 2\pi \sqrt{\frac{m}{k}} \); where m = M + m T = \( 2\pi \sqrt{\frac{M+m}{k}} \) (

B) Let block is displaced by x in (+ve) direction so force on block will be in (-ve) direction F = Kx (M + m)a = -Kx a = \( \frac{-Kx}{M+m} \) 

(C) As upper block is moving due to friction thus f = ma = \( \frac{mKx}{M+m} \) 

(D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together. 

KA = (M + m)a a = \( \frac{KA}{M+m} \) f = ma = \( \frac{mKA}{M+m} \) \( f_{max} = f_L = \mu mg \) f = \( \mu mg \) \( \frac{mKA}{M+m} = \mu mg \) A = \( \frac{\mu (M+m)g}{K} \) 

Maximum friction can be \( \mu mg \) as force is acting between blocks and normal force here is mg.