13.6 Energy in simple harmonic motion

Energy in SHM

The particle executing SHM has both types of energy: potential energy and kinetic energy. When a body is displaced from its equilibrium position by doing work upon it, the body acquires potential energy. However, when the body is released, it begins to move back with a velocity, thus acquiring kinetic energy.

Kinetic energy

The relation for kinetic energy of a particle executing SHM can be expressed in two forms, i.e. in terms of displacement and in terms of time. Let us consider
\(
\begin{aligned}
& x=A \sin (\omega t+\phi) \\
& v=A \omega \cos (\omega t+\phi)=\omega \sqrt{A^2-x^2}
\end{aligned}
\)
\(
\begin{aligned}
& \text { KE}=\frac{1}{2} m v^2 \quad\left[\text { Since, } v^2=A^2 \omega^2 \cos ^2(\omega t+\phi)\right] \\
& =\frac{1}{2} m \omega^2 A^2 \cos ^2(\omega t+\phi) \\
& =\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\
& \therefore \text { KE }=\frac{1}{2} m \omega^2 A^2 \cos ^2(\omega t+\phi)=\frac{1}{2} m \omega^2\left(A^2-x^2\right)
\end{aligned}
\)
where, \(A=\) amplitude of SHM.
As, \(\quad \omega=\sqrt{\frac{k}{m}} \Rightarrow m \omega^2=k \therefore {KE}=\frac{1}{2} k\left(A^2-x^2\right)\)

In terms of time Kinetic energy of a particle executing SHM at any time \(t\) is given by
\(
KE=\frac{1}{2} m \omega^2 A^2 \cos ^2 \omega t
\)
i.e. kinetic energy of a particle executing SHM is a periodic function of time.
The kinetic energy versus time graph is given below:

From the graph above, it is clear that at \(t=0, T / 2\), \(T\), etc., kinetic energy has its maximum value. For \(x= \pm A\) or \(t=\frac{T}{4}, \frac{3 T}{4}\), the kinetic energy has its minimum value.

Special cases:

• Kinetic energy is maximum at the mean position, i.e. at \(x=0\)
  \(
  \mathrm{KE}_{\max }=\frac{1}{2} m \omega^2 A^2
  \)
• Kinetic energy is minimum (zero) at the extreme position, i.e. at \(x= \pm A, \mathrm{KE}_{\min }=\) zero

Potential energy

Work done by the restoring force while displacing the particle from the mean position \((x=0)\) to \(x=x\) :

The work done by restoring force when the particle has been displaced from the position \(x\) to \({x}+{dx}\) is given by
\(
d w=F d x=-k x~ d x
\)
\(
\begin{aligned}
& w=\int d w=\int_0^x-k x~ d x=\frac{-k x^2}{2} \\
& =-\frac{m \omega^2 x^2}{2} \quad \left[k=m \omega^2\right]
\end{aligned}
\)
\(
=-\frac{m \omega^2}{2} A^2 \sin ^2(\omega t+\phi)
\)
Potential Energy = -(work done by restoring force)
Potential Energy \(U=\frac{m \omega^2 x^2}{2}=\frac{1}{2} k x^2=\frac{m \omega^2 A^2}{2} \sin ^2(\omega t+\phi)\)

Special cases:

• Potential energy is maximum at the extreme position, i.e. at \(x= \pm A\)
  \(
  U_{\max }=\frac{1}{2} m \omega^2 A^2
  \)
  Potential energy is minimum (zero) at the mean position, i.e. at \(x=0, U_{\min }=0\)
• In terms of time Potential energy of a particle executing SHM at any time \(t\) is given by
  \(
  U=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
  \)
  The potential energy versus time graph is given below

From the graph, it is clear that at mean position (i.e. \(x=0\) or \(t=0, T / 2, T \ldots)\), potential energy is minimum and at extreme position (i.e. \(\mathrm{t} x= \pm A\) or \(t=T / 4,3 T / 4\) ), its value is maximum. Clearly, twice in each cycle, potential energy acquires its peak values.
Hence, the potential energy is the periodic function of time with a period of \(T / 2\).

Total Mechanical Energy of the Particle Executing Simple Harmonic Motion

The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude.
\(
\begin{gathered}
E=KE+U=\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2 \\
E=\frac{1}{2} m \omega^2 A^2 \text { or } E=\frac{1}{2} k A^2
\end{gathered}
\)
which is a constant quantity, i.e. it remains the same at all instants and at all displacement.

The figure below shows the variation of kinetic energy \((K)\), potential energy \((U)\), and total energy \((E)\) as a function of time [shown in Figure (a)] and displacement [shown in Figure (b)] for a particle executing SHM.

\(\therefore\) From the above graph, it can be concluded that

• The total energy remains constant at all \(t\) or \(x\). So, it is shown by a straight line parallel to time.
• Also, \(U=\frac{1}{2} k x^2\) and \(K=\frac{1}{2} k\left(A^2-x^2\right)\), so the graphs of \(U\) and \(K\) versus \(x\) are parabolic.
• Both kinetic energy and potential energy reach maximum value twice, during each period of SHM.
• In the course of motion, kinetic energy increases at the expense of potential energy or vice-versa. Therefore, we may say that during an oscillation, there is a continuous exchange of kinetic and potential energies.

Important points regarding energy in SHM

• The frequency of oscillation of potential energy and kinetic energy is twice as that of displacement or velocity or acceleration.
• The average value of kinetic energy and potential energy is
  \(
  \begin{aligned}
  KE_{\mathrm{av}} & =U_{\mathrm{av}} \\
  & =\frac{1}{2} E=\frac{1}{4} k A^2
  \end{aligned}
  \)

Example 1: The potential energy of a particle executing SHM is 2.5 J when its displacement is half of the amplitude, then determine the total energy of the particle.

Solution: As, potential energy \(=\frac{1}{2} k x^2\)
\(
\Rightarrow \quad \frac{1}{2} k \frac{A^2}{4}=2.5 \mathrm{~J} \quad\left(\because x=\frac{A}{2}\right)
\)
where, \(x\) is the displacement of the particle and \(A\) is the amplitude.
\(
\Rightarrow \quad \text { Total energy }=\frac{1}{2} k A^2=2.5 \times 4=10 \mathrm{~J}
\)

Example 2: A harmonic oscillator of force constant \(4 \times 10^6 \mathrm{Nm}^{-1}\) and amplitude 0.01 m has total energy 240 J . What is maximum kinetic energy and minimum potential energy?

Solution: Given, \(k=4 \times 10^6 \mathrm{~N} / \mathrm{m}, A=0.01 \mathrm{~m}\), total energy \(=240 \mathrm{~J}\) Maximum kinetic energy
\(
=\frac{1}{2} m \omega^2 A^2=\frac{1}{2} k A^2=\frac{1}{2} \times 4 \times 10^6 \times(0.01)^2=200 \mathrm{~J}
\)
Minimum potential energy \(=\) Total energy – Maximum kinetic energy \(=40 \mathrm{~J}\)

Example 3: A linear harmonic oscillator has a total mechanical energy of 200 J . Potential energy of it at mean position is 50 J. Find
(i) the minimum potential energy,
(ii) the maximum kinetic energy,
(iii) the potential energy at extreme positions.

Solution: (i) At mean position, potential energy is minimum. Hence, \(\quad U_{\min }=50 \mathrm{~J}\)
(ii) At the mean position, kinetic energy is maximum.
\(
\therefore \quad K_{\max }=E-U_{\min }=200-50=150 \mathrm{~J}
\)
(iii) At extreme positions, kinetic energy is zero and potential energy is maximum.
\(
\therefore \quad U_{\max }=E=200 \mathrm{~J} \text { (From law of conservation of energy) }
\)

Example 4: A particle executes SHM, at what value of displacement are the kinetic and potential energies equal?

Solution: We know that, kinetic energy \(=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\) and potential energy \(U=\frac{1}{2} m \omega^2 x^2\)
Since, \(KE=U\)
\(
\Rightarrow \frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2 x^2 \text { or } 2 x^2=A^2
\)
\(
x= \pm \frac{A}{\sqrt{2}}
\)

Example 5: A particle starts oscillating simple harmonically from its equilibrium position with time period \(T\). Determine the ratio of kinetic energy and potential energy of the particle at time \(t=\frac{T}{12}\).

Solution: At \(t=\frac{T}{12}, x=A \sin \frac{2 \pi}{T} \times \frac{T}{12}=A \sin \frac{\pi}{6}=\frac{A}{2}\)
So, kinetic energy, \(\mathrm{KE}=\frac{1}{2} k\left(A^2-x^2\right)=\frac{3}{4} \times \frac{1}{2} k A^2\) and potential energy, \(\mathrm{PE}=\frac{1}{2} k x^2=\frac{1}{4} \times \frac{1}{2} k A^2\)
\(
\therefore \quad \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{3}{1}
\)

Example 6: A particle of mass 0.2 kg is executing \(S H M\) of amplitude 0.2 m. When it passes through the mean position, its kinetic energy is \(64 \times 10^{-3} \mathrm{~J}\). Obtain the equation of motion of this particle, if the initial phase of oscillation is \(\pi / 4\).

Solution: Here, \(A=0.2 \mathrm{~m}, \phi=\frac{\pi}{4}, \mathrm{KE}=64 \times 10^{-3} \mathrm{~J}\)
As we know that, \(\mathrm{KE}=\frac{1}{2} m \omega^2 A^2\) where, \(m=\) mass of particle \(=0.2 \mathrm{~kg}\)
\(
\begin{array}{ll}
\Rightarrow & 64 \times 10^{-3}=\frac{1}{2} \times 0.2 \times \omega^2 \times(0.2)^2 \\
\Rightarrow & \omega^2=\frac{128 \times 10^{-3}}{0.2 \times 0.2 \times 0.2} \text { or } \omega=4 \mathrm{~s}^{-1}
\end{array}
\)
\(\therefore\) Equation of motion can be written as
\(
x=A \sin (\omega t+\phi)=0.2 \sin \left(4 t+\frac{\pi}{4}\right)
\)

Example 7: A particle executes SHM with amplitude A and time period T. When the displacement from the equilibrium position is half the amplitude, what fractions of the total energy are kinetic and potential?

Solution: Total energy, \(E=\frac{1}{2} k A^2\)
When
\(
x= \pm \frac{A}{2} \text {, then }
\)
Potential energy, \(U=\frac{1}{2} k x^2=\frac{1}{2} k \cdot\left( \pm \frac{A}{2}\right)^2\)
\(
=\frac{1}{4} \cdot \frac{1}{2} k A^2=\frac{1}{4} E=25 \% \text { of } E
\)
\(\therefore\) Kinetic energy, \(E-U=E-\frac{1}{4} E=\frac{3}{4} E=75 \%\) of \(E\)

Example 8: The potential energy of a particle oscillating on \(X\)-axis is given as, \(U=20+(x-2)^2\) Here, \(U\) is in joule and \(x\) is in metre. The total mechanical energy of the particle is 36 J.
(i) State whether the motion of the particle is simple harmonic or not.
(ii) Find the mean position.
(iii) Find the maximum kinetic energy of the particle.

Solution: (i) \(F=-\frac{d U}{d x}=-2(x-2)\)
By assuming \(x-2=X\), we get
Since,
\(
\begin{aligned}
& F=-2 X \\
& F \propto-X
\end{aligned}
\)
The motion of the particle is simple harmonic.
(ii) The mean position of the particle is \(X=0\) or \(x-2=0\), which gives \(x=2 \mathrm{~m}\)
(iii) At \(x=2 \mathrm{~m}\) (mean position),
\(
U_{\min }=20+(2-2)^2=20 \mathrm{~J}
\)
\(\therefore \quad\) Maximum kinetic energy of the particle,
\(
K_{\max }=E-U_{\min }=36-20=16 \mathrm{~J}
\)
Note: \(U_{\text {min }}\) is 20 J at mean position or at \(x=2 \mathrm{~m}\).

Example 9: A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of \(50 \mathrm{~N} \mathrm{~m}^{-1}\). The block is pulled to a distance \(x=10 \mathrm{~cm}\) from its equilibrium position at \(x=0\) on a frictionless surface from rest at \(t=0\). Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Solution: The block executes SHM, its angular frequency, as given by
\(
\begin{aligned}
\omega & =\sqrt{\frac{k}{m}} \\
& =\sqrt{\frac{50 \mathrm{~N} \mathrm{~m}^{-1}}{1 \mathrm{~kg}}} \\
& =7.07 \mathrm{rad} \mathrm{~s}^{-1}
\end{aligned}
\)
Its displacement at any time t is then given by,
\(
x(t)=0.1 \cos (7.07 t)
\)
Therefore, when the particle is 5 cm away from the mean position, we have
\(
0.05=0.1 \cos (7.07 t)
\)
Or \(\cos (7.07 t)=0.5\) and hence
\(
\sin (7.07 t)=\frac{\sqrt{3}}{2}=0.866
\)
Then, the velocity of the block at \(x=5 \mathrm{~cm}\) is
\(
\begin{aligned}
& =0.1 \times 7.07 \quad 0.866 \mathrm{~m} \mathrm{~s}^{-1} \\
& =0.61 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
Hence the K.E. of the block,
\(
\begin{aligned}
= & \frac{1}{2} m v^2 \\
& =1 / 2\left[1 \mathrm{~kg} \times\left(0.6123 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\right] \\
& =0.19 \mathrm{~J}
\end{aligned}
\)
The P.E. of the block,
\(
\begin{aligned}
& =\frac{1}{2} k x^2 \\
& =1 / 2\left(50 \mathrm{~N} \mathrm{~m}^{-1} \times 0.05 \mathrm{~m} \times 0.05 \mathrm{~m}\right) \\
& =0.0625 \mathrm{~J}
\end{aligned}
\)
The total energy of the block at \(x=5 \mathrm{~cm}\),
\(
\begin{aligned}
& =\text { K.E. + P.E. } \\
& =0.25 \mathrm{~J}
\end{aligned}
\)
we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,
\(
\begin{aligned}
& =1 / 2\left(50 \mathrm{~N} \mathrm{~m}^{-1} \times 0.1 \mathrm{~m} \times 0.1 \mathrm{~m}\right) \\
& =0.25 \mathrm{~J}
\end{aligned}
\)
which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of conservation of energy.