Two blocks of masses $m$ and $M$ $(M > m)$ are placed on a frictionless table as shown in the figure. $A$ massless spring with spring constant $k$ is attached to 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 $f = \frac{mkx}{M + m}$
$(D)$ The maximum amplitude of the upper block,if it does not slip,is $A = \frac{\mu mg(M + m)}{mk} = \frac{\mu g(M + m)}{k}$ (Wait,let's re-evaluate: $f_{max} = \mu mg$. Since $f = ma = m \cdot \frac{kx}{M+m}$,at max amplitude $A$,$m \cdot \frac{kA}{M+m} = \mu mg \implies A = \frac{\mu g(M+m)}{k}$)
$(E)$ Maximum frictional force can be $\mu mg$.
Choose the correct answer from the options given below.

  • A
    $A, B, C, E$ Only
  • B
    $B, C, D$ Only
  • C
    $A, B, C, D$ Only
  • D
    $A, B, C$ Only

Explore More

Similar Questions

$A$ body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The mass of the spring is negligible. When the mass $m$ is pulled down slightly and released,it oscillates with a time period of $3 \ s$. When the mass $m$ is increased by $1 \ kg$,the time period of oscillation becomes $5 \ s$. What is the value of $m$ in $kg$?

$A$ $5\, kg$ collar is attached to a spring of spring constant $500\, N/m$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $10\, cm$ and released. The time period of oscillation is

$A$ $3 \ kg$ block is connected as shown in the figure. The spring constants of the two springs $K_1$ and $K_2$ are $50 \ Nm^{-1}$ and $150 \ Nm^{-1}$ respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is $(g=10 \ ms^{-2})$. (in $ms^{-2}$)

$A$ body of mass $1 \,kg$ is attached to the lower end of a vertically suspended spring of force constant $600 \,N \,m^{-1}$. If another body of mass $0.5 \,kg$ moving vertically upward hits the suspended body with a velocity $3 \,m \,s^{-1}$ and gets embedded in it, then the frequency of the oscillation is

$A$ block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 \, cm$ from its equilibrium position (at $x = 0$) on a frictionless surface from rest. The total energy of the block at $x = 5 \, cm$ is $0.25 \, J$. The spring constant of the spring is $......... \, N \, m^{-1}$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo