
The potential energy of a particle with displacement x is $U\left( x \right)$ . The motion is simple harmonic. If K is a positive constant then
(A) $U = kx$
(B) $U = k$
(C) $U = - k{x^2}/2$
(D) $U = k{x^2}$
Answer
162.6k+ views
Hint:
First start with finding the relation of the potential 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 and finally get the right answer and 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 : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Potential energy is denoted as U(x) in the question.
So, $U\left( x \right) = \dfrac{1}{2}m{\omega ^2}{x^2}$
Where $K = \dfrac{1}{2}m{\omega ^2}$
So, $U\left( x \right) = K{x^2}$
Hence the correct answer is Option(D).
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. Use the process in sequence then only you will get the right answer for the given question.
First start with finding the relation of the potential 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 and finally get the right answer and 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 : $P.E. = 1/2{\text{ }}k{x^2}\; = {\text{ }}1/2\;m\;{\omega ^2}{x^2}$
Potential energy is denoted as U(x) in the question.
So, $U\left( x \right) = \dfrac{1}{2}m{\omega ^2}{x^2}$
Where $K = \dfrac{1}{2}m{\omega ^2}$
So, $U\left( x \right) = K{x^2}$
Hence the correct answer is Option(D).
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. Use the process in sequence then only you will get the right answer for the given question.
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