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Physics · Simple Harmonic Motion

JEE Main 2025 — 3 April, Morning Shift — Question 59

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

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

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

C. The magnitude of the frictional force on the upper block is mμ∣x∣M+m\frac{m \mu|x|}{M+m}

D. The maximum amplitude of the upper block, if it does not slip, is μ(M+m)gk\frac{\mu(M+m) g}{k} E. Maximum frictional force can be μ(M+m)g\mu(M+m) g Choose the correct answer from the options given below :

Question figure
  1. Option A:

    B, C, D only

  2. Option B:

    A, B, C only

  3. Option C:

    A, B, D only

    Correct
  4. Option D:

    C, D, E only

Answer: C

Step-by-step solution

A. Assuming no slipping, T=2πmtotal kT=2 \pi \sqrt{\frac{m_{\text {total }}}{k}} A is correct.

B. Assuming no slipping, a=∣F∣ma=\frac{|F|}{m} BB is correct.

C. f=(m)(a)=m×kxm+Mf=(m)(a)=\frac{m \times k x}{m+M} C is correct.

D. For no slipping kx0m+M≤μg\frac{k x_{0}}{m+M} \leq \mu g

DD is correct. E. fmax =μmgf_{\text {max }}=\mu m g EE is incorrect.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Physics
Chapter
Simple Harmonic Motion
Topic
Miscellaneous Problems in SHM
Two blocks of masses m and M,(M m) are placed on a frictionless table… | JEE Main 2025 PYQ with Solution · DhiX AI