Question

The amplitude of a particle in SHM is 0.2 m. The P.E. and K.E. of this particle 0.1 sec after passing the mean position are equal. The period of oscillation is:
(A) 1 s
(B) 0.8 s
(C) 0.6 s
(D) 0.4 s

Answer 1

Hint To answer this question it is required to form the expressions for the potential energy and the expression for the kinetic energy. Then we have to equate them and solve to get the required answer for this question.

Complete step by step answer
We know that the amplitude of the particle in SHM is given as A which is 0.2 m.
The potential energy is given as: $2\pi$ $\dfrac{1}{2}m{\omega ^2}{x^2}$
The kinetic energy is given as: $\dfrac{1}{2}m{\omega ^2}({A^2} - {x^2})$
Now we can write that Potential Energy = Kinetic Energy
So, putting the values in the above relation we get that:
$\dfrac{1}{2}m{\omega ^2}{x^2} - \dfrac{1}{2}m{\omega ^2}({A^2} - {x^2})$
So, we can write that:
${x^2} = {A^2} - {x^2}$
$\Rightarrow x = \dfrac{A}{{\sqrt 2 }}$
So now we can write that
$\sin \theta = \dfrac{{\dfrac{A}{{\sqrt 2 }}}}{A}$
$\Rightarrow \theta = \dfrac{\pi }{4}{\text{ }}$
Therefore, for 0.1 sec for $\dfrac{\pi }{4}$
Therefore, for $2\pi$time taken to 0.8 sec
Therefore, the time period is 0.8 sec.
So, the period of oscillation is 0.8 secs.

Hence the correct answer is option B.

Note It should be known to us that period of oscillation is defined as the smallest possible time interval. In this small duration a system can undergo oscillation and at the same time can return to the state it was in at a specific time arbitrarily chosen, at the very initial point of oscillation.