
The potential energy of a particle executing simple harmonic motion (S.H.M.) is $2.5J$ , when its displacement is half of the amplitude. The total energy of the particle will be:
(A) $18J$
(B) $10J$
(C) $12J$
(D) $2.5J$
Answer
161.7k+ views
Hint:
First start with finding the relation of the potential energy and the total energy of a particle executing simple harmonic motion (S.H.M.) and try to find out which of the given options is fit in that relation, you can use the method of elimination and can eliminate the wrong option one by one.
Formula used :
Potential energy : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Complete step by step solution:
Potential energy is the energy possessed by the particle when the particle is at rest.
Now when the particle is executing simple harmonic motion at a distance x from the mean position.
The force acting will be $F = - kx$
Now the work done will be:
$dW = - fdx$
After solving, we get;
Total work done, $W = 1/2{\text{ }}K{\text{ }}{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
The total work done here will get stored in the form of potential energy.
So, Potential energy : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Now, total energy is T.E. = Kinetic energy + potential energy
$T.E = 1/2{\text{ }}K{\text{ }}{a^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{a^2}$
$
\dfrac{U}{{T.E}} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{x^2}}}{{\dfrac{1}{2}m{\omega ^2}{a^2}}} \\
\dfrac{U}{{T.E}} = \dfrac{{{x^2}}}{{{a^2}}} \\
\dfrac{U}{{T.E}} = \dfrac{{{{(\dfrac{a}{2})}^2}}}{{{a^2}}} \\
$ (since $x$ is half of the amplitude)
$ \Rightarrow \dfrac{{2.5}}{{T.E}} = \dfrac{1}{4}$
By solving;
$T.E = 10J$
Hence the correct answer is Option(B).
Note:
First find the force then the displacement of the particle in the simple harmonic motion and then use the value of the force in finding the work done and finally get the potential energy. Then total energy is equal to sum of kinetic and potential energy but the kinetic energy here is zero.
First start with finding the relation of the potential energy and the total energy of a particle executing simple harmonic motion (S.H.M.) and try to find out which of the given options is fit in that relation, you can use the method of elimination and can eliminate the wrong option one by one.
Formula used :
Potential energy : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Complete step by step solution:
Potential energy is the energy possessed by the particle when the particle is at rest.
Now when the particle is executing simple harmonic motion at a distance x from the mean position.
The force acting will be $F = - kx$
Now the work done will be:
$dW = - fdx$
After solving, we get;
Total work done, $W = 1/2{\text{ }}K{\text{ }}{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
The total work done here will get stored in the form of potential energy.
So, Potential energy : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Now, total energy is T.E. = Kinetic energy + potential energy
$T.E = 1/2{\text{ }}K{\text{ }}{a^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{a^2}$
$
\dfrac{U}{{T.E}} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{x^2}}}{{\dfrac{1}{2}m{\omega ^2}{a^2}}} \\
\dfrac{U}{{T.E}} = \dfrac{{{x^2}}}{{{a^2}}} \\
\dfrac{U}{{T.E}} = \dfrac{{{{(\dfrac{a}{2})}^2}}}{{{a^2}}} \\
$ (since $x$ is half of the amplitude)
$ \Rightarrow \dfrac{{2.5}}{{T.E}} = \dfrac{1}{4}$
By solving;
$T.E = 10J$
Hence the correct answer is Option(B).
Note:
First find the force then the displacement of the particle in the simple harmonic motion and then use the value of the force in finding the work done and finally get the potential energy. Then total energy is equal to sum of kinetic and potential energy but the kinetic energy here is zero.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
