Question

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35. A system is shown in figure. Find the time period
for small oscillations of two blocks.
k
2k
m
m
(1) 210
21,
3m
K k
(2) 2r,
3m
2k
3m
(3) 24K
(4) 2r,
3m
8k​

Answer 1

Answer:

Let the mass m be displaced by a small distance x to the right from its mean position as shown in figure (b). Due to it the spring on the left side gets stretched by a length x while that on the right side gets compressed by the same length. The forces acting on the mass are

F1 = -kx towards left hand side

F2 = -kx towards left hand side

The net force acting on the mass is, F=F1+F2=−2kx.

Here F∝x and -ve sign shows that force is towards the mean position, therefore the motion executed by the particle is simple harmonic.

Its acceleration is

a=mF=−m2kx      ....(i)

The standard equation of SHM is

a=−ω2x ...(ii)

Comparing (i) and (ii), we get

ω2=m2korω=m2k

Time period, T=ω2π=2π2km