Question

When the displacement in SHM is one-half the amplitude${x_m}$, what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

Answer 1

We are going to firstly take the displacement for the Simple harmonic motion and find corresponding velocity. For that, the potential and the kinetic energy are calculated at the half displacement and also their ratios with total energy is calculated. Also the displacement is found.

Formula used: The potential and kinetic energies are given by:

$$U\left( t \right) = \dfrac{1}{2}k{x^2}\left( t \right) \\\ K\left( t \right) = \dfrac{1}{2}m{v^2}\left( t \right) \\\$$

$x(t)$is the displacement and $v\left( t \right)$is the velocity.

Complete step-by-step solution:
It is given that the displacement in SHM is:
$x\left( t \right) = \dfrac{{{x_m}}}{2}$
Where ${x_m}$is the amplitude
The displacement in Simple Harmonic Motion is given by:
$x(t) = {x_m}\cos \left( {\omega t + \phi } \right) - - - - - \left( 1 \right)$
And the velocity is given by
$v\left( t \right) = - \omega {x_m}\sin \left( {\omega t + \phi } \right) - - - - - \left( 2 \right)$
The potential and kinetic energies are given by:

$$U\left( t \right) = \dfrac{1}{2}k{x^2}\left( t \right) \\\ K\left( t \right) = \dfrac{1}{2}m{v^2}\left( t \right) \\\$$

Substituting $\left( 1 \right)$and$\left( 2 \right)$, in the above two equations.

$$U\left( t \right) = \dfrac{1}{2}k{x_m}^2{\cos ^2}\left( {\omega t + \phi } \right) - - - - - \left( 3 \right) \\\ K\left( t \right) = \dfrac{1}{2}m{\omega ^2}{x_m}^2{\sin ^2}\left( {\omega t + \phi } \right) = \dfrac{1}{2}k{x_m}^2{\sin ^2}\left( {\omega t + \phi } \right) - - - - - \left( 4 \right) \\\$$

Total energy is given by:
$E = U\left( t \right) + K\left( t \right) = \dfrac{1}{2}k{x_m}^2\left[ {{{\cos }^2}\left( {\omega t + \phi } \right) + {{\sin }^2}\left( {\omega t + \phi } \right)} \right] = \dfrac{1}{2}k{x_m}^2 - - - - - \left( 5 \right)$
Now, when the displacement is one half the amplitude${x_m}$, i.e. $x\left( t \right) = \dfrac{{{x_m}}}{2}$
From$\left( 1 \right)$, $\cos \left( {\omega t + \phi } \right) = \dfrac{1}{2}$
Since, ${\cos ^2}\left( {\omega t + \phi } \right) + {\sin ^2}\left( {\omega t + \phi } \right) = 1$,
This implies, $\sin \left( {\omega t + \phi } \right) = \sqrt {1 - {{\cos }^2}\left( {\omega t + \phi } \right)} = \sqrt {1 - \dfrac{1}{4}} = \dfrac{{\sqrt 3 }}{2}$
$\left( a \right)$Dividing $\left( 4 \right)$and$\left( 5 \right)$, and substitute with these values, we get,

$$\dfrac{K}{E} = {\sin ^2}\left( {\omega t + \phi } \right) = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = \dfrac{3}{4} \\\ \Rightarrow \dfrac{K}{E} = \dfrac{3}{4} \\\$$

$\left( b \right)$Dividing $\left( 3 \right)$and$\left( 5 \right)$, substituting these values, we get

$$\dfrac{U}{E} = {\cos ^2}\left( {\omega t + \phi } \right) = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4} \\\ \Rightarrow \dfrac{U}{E} = \dfrac{1}{4} \\\$$

$\left( c \right)$Finally, we need to find at what displacement is the energy of the system half kinetic energy and half potential energy, i.e.,
$\dfrac{U}{E} = \dfrac{K}{E} = \dfrac{1}{2}$
Dividing $U = \dfrac{1}{2}k{x^2}\left( t \right)$by$E = \dfrac{1}{2}k{x^2}_m$, we get

$$\dfrac{U}{E} = \dfrac{{{x^2}}}{{{x^2}_m}} \\\ \Rightarrow \dfrac{1}{2} = \dfrac{{{x^2}}}{{{x^2}_m}} \\\$$

Solving for$x$, we get
$x = \dfrac{{{x_m}}}{{\sqrt 2 }}$
Therefore,

$$\left( a \right)\dfrac{K}{E} = \dfrac{3}{4} \\\ \left( b \right)\dfrac{U}{E} = \dfrac{1}{4} \\\ \left( c \right)x = \dfrac{{{x_m}}}{{\sqrt 2 }} \\\$$

Note: It is important to note that simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. The proportion of kinetic energy to the total energy is more.