
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
497.4k+ views
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

