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In the arrangement, spring constant k has value 2Nm1, mass M=3kg and mass M=1kg. Mass M is in contact with a smooth surface. The coefficient of friction between two blocks is 0.1. The time period of SHM executed by the system is …

A) π6
B) π2
C) 22π
D) 2π

Answer
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Hint: For small amplitude the two blocks oscillate together , with angular frequency ω=kM+mhz. So the time period becomes T=2πMk.

Formula used:- T=2πMk where, T = Time- period of the spring block system
                                                   M = Total mass attached to spring
Value of the spring constant is

                                                    k = spring constant of the spring
Since total mass is the sum of both the blocks.
M=M+ m = 4kg
k= 2Nm1
We have to find the time period of the systemT=2πMk
Putting the valueM= 4kg,& k= 2Nm1in order to find the time period
T=2π42=22πsec.

Hence option (C ) is the correct option.

Note :- Periodic motion : Any motion which repeats itself after a regular interval of time is called the periodic motion. Few examples are motion of planets around the sun , motion of the pendulum of wall clock.
-Oscillatory motion : The motion of a body is oscillatory if it moves back and forth ( to and fro) about a fixed point after a regular interval of time. This fixed position is called the mean position or the equilibrium position of the body.
-The periodic time of a hard spring is less as compared to that of a sift spring because the spring constant is large for a hard spring.
-For a system executing SHM, the mechanical energy which is the sum of potential energy and kinetic energy together remains constant.
-The frequency of oscillation of potential energy and kinetic energy is twice as of the displacement or velocity or acceleration of a particle executing simple harmonic motion.

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