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Velocity Displacement Relationship

In the arrang...

Question

In the arrangement shown, both the springs are in their natural lengths. The coefficient of friction between $m_{2}$ and $m_{1}$ is $μ$ . There is no friction between $m_{1}$ and surface. If the blocks are displaced slightly, they perform SHM together.
then, the
(1) frequency of such oscillations $\frac{1}{2 π} \sqrt{\frac{k_{1} + k_{2}}{m_{1} + m_{2}}},$
(2) condition if the frictional force on block $m_{2}$ is to act in the direction of its displacement from the mean position is $\frac{m_{1}}{m_{2}} > \frac{k_{1}}{k_{2}}$
(3) If the condition obtained in (b) is met, the maximum amplitude of their oscillations is $\frac{μ m_{2} g (m_{1} + m_{2})}{k_{2} m_{1} - k_{1} m_{2}}$

A

1, 2 are correct

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B

2, 3 are correct

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C

all three are correct

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D

1 is only correct

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Solution

The correct option is C all three are correct
(a) If the blocks oscillate together, the equivalent spring constant is $(k_{1} + k_{2})$
$∴$ Frequency = $\frac{1}{2 π} \sqrt{\frac{k_{1} + k_{2}}{m_{1} + m_{2}}}$
(b) Assuming that the blocks $(m_{1} + m_{2})$ are displaced towards right, the various forces (except friction) on $m_{2}$ in the reference frame of m1 are as shown. where a = $(\frac{k_{1} + k_{2}}{m_{1} + m_{2}}) x$
For friction to act on $m_{2}$ in the direction of its displacement,
$k_{2} x > m_{2} (\frac{k_{1} + k_{2}}{m_{1} + m_{2}}) x$
$\Rightarrow m_{1} k_{2} + m_{2} k_{2} - m_{2} k_{1} - m_{2} k_{2} > 0$
$\Rightarrow m_{1} k_{2} - m_{2} k_{1} > 0$
$\Rightarrow \frac{m_{1}}{m_{2}} > \frac{k_{1}}{k_{2}}$ This is the desired condition.
(c) Assuming , $\frac{m_{1}}{m_{2}} > \frac{k_{1}}{k_{2}}$ the free body diagram of $m_{2}$ with $m_{1}$ and $m_{2}$ together displaced towards right is as shown.
If A is amplitude of oscillation of $m_{2}, k_{2} A = f + m_{2} ω^{2} A$
$A = \frac{μ m_{2} g}{k_{2} - m_{2} ω^{2}} = \frac{μ m_{2} g}{k_{2} - m_{2} (\frac{k_{1} + k_{2}}{m_{1} + m_{2}})}$
$= \frac{μ m_{2} g (m_{1} + m_{2})}{k_{2} m_{1} - k_{1} m_{2}}$

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