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Question: One end of a spring of force constant \(k\) is fixed to a vertical wall and another to a body of mas...

One end of a spring of force constant kk is fixed to a vertical wall and another to a body of mass mm resting on a smooth horizontal surface. There is another wall at the distance x0{x_0} from the body. The spring is then compressed by 3x03{x_0} and released.The time taken to strike the wall from the instant of release is ? (given sin1(13)=(π9){\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \left( {\dfrac{\pi }{9}} \right))

A. π6mk\dfrac{\pi }{6}\sqrt {\dfrac{m}{k}}
B. 2π3mk\dfrac{{2\pi }}{3}\sqrt {\dfrac{m}{k}}
C. π4mk\dfrac{\pi }{4}\sqrt {\dfrac{m}{k}}
D. 11π9mk\dfrac{{11\pi }}{9}\sqrt {\dfrac{m}{k}}

Explanation

Solution

In this question, we have to calculate the time taken by the mass to strike from the instant of release. This can be achieved by using the concept of displacement of a body from mean position and time needed to cover the distance from compressed position to mean position.

Complete step by step answer:
The total amplitude of the motion is A=2x0A = 2{x_0}. The time needed to cover from compressed position to mean position is T4\dfrac{T}{4}, where TT is the total time required to do the motion.

Now, the displacement from the mean position to x0{x_0} , time taken t will be given as
y=Asinwty = A\sin wt
x0=2x0sinwt\Rightarrow {x_0} = 2{x_0}\sin wt
But w=2πTw = \dfrac{{2\pi }}{T}, So,
x0=2x0sin2πTt\Rightarrow {x_0} = 2{x_0}\sin \dfrac{{2\pi }}{T}t
solving for the value of tt, we get t=T12(1)t = \dfrac{T}{{12}} - - - - - - (1)
Thus, Time taken to hit the wall is T12+T4=T3(2)\dfrac{T}{{12}} + \dfrac{T}{4} = \dfrac{T}{3} - - - - - - (2)
Now, If mass mm is suspended from the spring of force constant KK , time period is given as
T=2πmkT = 2\pi \sqrt {\dfrac{m}{k}}
So, Time taken to hit the wall is T3=2π3mk\dfrac{T}{3} = \dfrac{{2\pi }}{3}\sqrt {\dfrac{m}{k}} .

Hence, option B is correct.

Note: The particle performing SHM starting from mean position is given by x=asinwtx = a\sin wt. The period of SHM does not depend on amplitude or energy of the particle. When the spring is compressed by 3x03{x_0} , the total path of the particle is divided into four equal parts of time T4\dfrac{T}{4}.