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Question

In the arrangement shown, both the springs are in their natural lengths. The coefficient of friction between m2 and m1 is μ. There is no friction between m1 and surface. If the blocks are displaced slightly, they perform SHM together.
then, the
(1) frequency of such oscillations 12πk1+k2m1+m2,
(2) condition if the frictional force on block m2 is to act in the direction of its displacement from the mean position is m1m2>k1k2
(3) If the condition obtained in (b) is met, the maximum amplitude of their oscillations is μm2g(m1+m2)k2m1k1m2

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 (k1+k2)
Frequency = 12πk1+k2m1+m2
(b) Assuming that the blocks (m1+m2) are displaced towards right, the various forces (except friction) on m2 in the reference frame of m1 are as shown. where a = (k1+k2m1+m2)x
For friction to act on m2 in the direction of its displacement,
k2x>m2(k1+k2m1+m2)x
m1k2+m2k2m2k1m2k2>0
m1k2m2k1>0
m1m2>k1k2 This is the desired condition.
(c) Assuming , m1m2>k1k2 the free body diagram of m2 with m1 and m2 together displaced towards right is as shown.
If A is amplitude of oscillation of m2,k2A=f+m2ω2A
A=μm2gk2m2ω2=μm2gk2m2(k1+k2m1+m2)
=μm2g(m1+m2)k2m1k1m2



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