
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
571.2k+ views
Hint: 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.
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.
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