In the arrangement shown both the springs are in their
![In the arrangement shown both the springs are in their](https://static.tllms.com/ckeditor_assets/pictures/14467/content_a.png)
then, the
(1) frequency of such oscillations 12π√k1+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)k2m1−k1m2
![](https://static.tllms.com/ckeditor_assets/pictures/14467/content_a.png)
A. 1, 2 are correct
B. 2, 3 are correct
C. all three are correct
D. only 1 is correct
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
(a) If the blocks oscillate together, the equivalent spring constant is (k1+k2)
∴ Frequency = 12π√k1+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+m2k2−m2k1−m2k2>0
⇒m1k2−m2k1>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=μm2gk2−m2ω2=μm2gk2−m2(k1+k2m1+m2)
=μm2g(m1+m2)k2m1−k1m2