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 = \( 2\pi \sqrt{\frac{M+m}{k}} \)
• 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 \( \frac{m\mu x}{M+m} \)
• D. The maximum amplitude of the upper block, if it does not slip, is \( \frac{\mu (M+m)g}{k} \)

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

(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.