Question

In the arrangement, spring constant $k$ has value $2N{m^{ - 1}}$, 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) $\pi \sqrt 6$
B) $\pi \sqrt 2$
C) $2\sqrt 2 \pi$
D) $2\pi$

Answer 1

For small amplitude the two blocks oscillate together , with angular frequency $\omega = \sqrt {\dfrac{k}{{M + m}}}$hz. So the time period becomes $T = 2\pi \sqrt {\dfrac{M}{k}}$.

Formula used:- $T = 2\pi \sqrt {\dfrac{M}{k}}$ 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 = $4kg$
$k$= $2N{m^{ - 1}}$
We have to find the time period of the system$T = 2\pi \sqrt {\dfrac{M}{k}}$
Putting the value$M$= $4kg$,& $k$= $2N{m^{ - 1}}$in order to find the time period
$T = 2\pi \sqrt {\dfrac{4}{2}}$=$2\sqrt 2 \pi$sec.

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.