A particle starts oscillating simple harmonically from its equilibrium position then the ratio of Kinetic energy and Potential energy of the particle at the same time $\dfrac{T}{{12}}$ (T is time period)
A. $2:1$
B. $3:1$
C. $4:1$
D. $1:4$
Answer
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Hint: In order to solve this question, we should know that when a body oscillate a simple harmonic motion it oscillate between its two extreme positions ad at extreme position it has completely Potential energy and at mean position it has completely Kinetic energy, here we will use the general equation of SHM and will calculate ratio of Kinetic energy and potential energy at given period of time.
Formula Used:
The position of a body performing SHM is written as
$x = A\sin \omega t$
where, $x,\omega ,A,t$ are position, angular velocity, amplitude, time of the body.
$K.E = \dfrac{1}{2}K({a^2} - {x^2})$
where K.E stands for Kinetic energy, K is constant.
$P.E = \dfrac{1}{2}K{x^2}$
where, P.E stands for Potential energy.
Complete step by step answer:
Let us find the value of position x at given time of $t = \dfrac{T}{{12}}$ and also we have, $\omega = \dfrac{{2\pi }}{T}$ we get,
$x = A\sin \omega t$
$\Rightarrow x = A\sin (\dfrac{{2\pi }}{T} \times \dfrac{T}{{12}})$
$\Rightarrow x = A\sin \dfrac{\pi }{6}$ since $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
$ \Rightarrow x = \dfrac{A}{2}$
Now put this value in formula of K.E and P.E we get,
$K.E = \dfrac{1}{2}K({a^2} - \dfrac{{{a^2}}}{4})$
$ \Rightarrow K.E = \dfrac{1}{2}K(\dfrac{{3{a^2}}}{4}) \to (i)$
and for P.E we have,
$P.E = \dfrac{1}{2}K(\dfrac{{{a^2}}}{4}) \to (ii)$
Divide the equation (i) by (ii) we get,
$\dfrac{{K.E}}{{P.E}} = \dfrac{{\dfrac{1}{2}K(\dfrac{{3{a^2}}}{4})}}{{\dfrac{1}{2}K(\dfrac{{{a^2}}}{4})}}$
$\therefore \dfrac{{K.E}}{{P.E}} = \dfrac{3}{1}$
so, the ratio of Kinetic energy to potential energy is $3:1$
Hence, the correct option is B.
Note:It should be remembered that, A Simple harmonic motion is one in which a body oscillate between two fixed points back and forth or up and down with a definite period of time which is called time period of the oscillation, also amplitude A is the maximum displacement of a body from its mean position to the extreme position.
Formula Used:
The position of a body performing SHM is written as
$x = A\sin \omega t$
where, $x,\omega ,A,t$ are position, angular velocity, amplitude, time of the body.
$K.E = \dfrac{1}{2}K({a^2} - {x^2})$
where K.E stands for Kinetic energy, K is constant.
$P.E = \dfrac{1}{2}K{x^2}$
where, P.E stands for Potential energy.
Complete step by step answer:
Let us find the value of position x at given time of $t = \dfrac{T}{{12}}$ and also we have, $\omega = \dfrac{{2\pi }}{T}$ we get,
$x = A\sin \omega t$
$\Rightarrow x = A\sin (\dfrac{{2\pi }}{T} \times \dfrac{T}{{12}})$
$\Rightarrow x = A\sin \dfrac{\pi }{6}$ since $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
$ \Rightarrow x = \dfrac{A}{2}$
Now put this value in formula of K.E and P.E we get,
$K.E = \dfrac{1}{2}K({a^2} - \dfrac{{{a^2}}}{4})$
$ \Rightarrow K.E = \dfrac{1}{2}K(\dfrac{{3{a^2}}}{4}) \to (i)$
and for P.E we have,
$P.E = \dfrac{1}{2}K(\dfrac{{{a^2}}}{4}) \to (ii)$
Divide the equation (i) by (ii) we get,
$\dfrac{{K.E}}{{P.E}} = \dfrac{{\dfrac{1}{2}K(\dfrac{{3{a^2}}}{4})}}{{\dfrac{1}{2}K(\dfrac{{{a^2}}}{4})}}$
$\therefore \dfrac{{K.E}}{{P.E}} = \dfrac{3}{1}$
so, the ratio of Kinetic energy to potential energy is $3:1$
Hence, the correct option is B.
Note:It should be remembered that, A Simple harmonic motion is one in which a body oscillate between two fixed points back and forth or up and down with a definite period of time which is called time period of the oscillation, also amplitude A is the maximum displacement of a body from its mean position to the extreme position.
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