Question

A spring-mass oscillator has a total energy \[{E_0}\] and amplitude \[{x_0}\]. For what value of \[x\] will \[K = U\]?
A. \[x = {x_0}\]
B. \[x = \dfrac{{{x_0}}}{{\sqrt2 }}\]
C. \[x = 2\]
D. \[x = 0\]

Answer 1

Hint: A spring mass oscillator is a form of simple harmonic motion (SHM). The total energy of SHM is equal to the maximum value of potential energy and maximum value of kinetic energy. Thus, the potential energy and kinetic energy will be equal to each when one considers the total energy of SHM.

Formula Used:
The maximum value of potential energy \[U\] and Kinetic energy \[K\] of a particle executing SHM is given by \[U = \dfrac{1}{2}k{x_0}^2\] and \[K = \dfrac{1}{2}k{x_0}^2\].

Complete step by step answer:
A motion that is oscillatory as well as periodic is called a Simple Harmonic Motion (SHM). In SHM, the acceleration of a particle is always proportional to the displacement from an equilibrium position.

Maximum displacement in SHM on either side of the mean position is called amplitude. The maximum value of potential energy \[U\] of a particle executing SHM and with an instantaneous displacement \[{x_0} = A\sin \omega t\] is given by,
\[U = \dfrac{1}{2}k{x_0}^2\]
where \[k\] is the force constant and \[y\] is the instantaneous displacement. The potential energy is maximum at the extreme position of oscillation and is minimum at the mean position of oscillation.

The maximum value of Kinetic energy is also given as \[K = \dfrac{1}{2}k{x_0}^2\]. The total energy of the system is given as the maximum value of Kinetic energy and the maximum value of Potential energy.
\[\therefore{E_0} = U = K = \dfrac{1}{2}k{x_0}^2\].
Thus, for the Kinetic energy and potential energy to be equal to each other, \[x\] must be equal to \[{x_0}\].

Hence, option A is the correct answer.

Note: If \[\omega \] is the amplitude and \[y\]is the instantaneous velocity, the potential energy \[U\]of a particle executing SHM can also be written as:
\[U = \dfrac{1}{2}m{\omega ^2}{y^2}\]
The minimum value of potential energy is zero, but the maximum value is given by:
\[U = \dfrac{1}{2}k{x_0}^2\]