14.7 Energy in simple harmonic motion

Both the kinetic and potential energies of a particle in SHM vary between zero and their maximum values.

we have seen that the velocity of a particle executing SHM is a periodic function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy \((K)\) of such a particle, which is defined as
\(
\begin{aligned}
K & =\frac{1}{2} m v^2 \\
& =\frac{1}{2} m \omega^2 A^2 \sin ^2(\omega t+\phi) \\
& =\frac{1}{2} k A^2 \sin ^2(\omega t+\phi) \dots(14.15)
\end{aligned}
\)
is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position. Note, since the \(\operatorname{sign}\) of \(v\) is immaterial in \(K\), the period of \(K\) is \(T / 2\).
The spring force \(F=-k x\) is a conservative force, with associated potential energy
\(
U=\frac{1}{2} k x^2 \dots(14.16)
\)
Hence the potential energy of a particle executing simple harmonic motion is,
\(
\begin{aligned}
& U(x)=\frac{1}{2} k x^2 \\
& =\frac{1}{2} k A^2 \cos ^2(\omega t+\phi) \dots(14.17)
\end{aligned}
\)
Thus, the potential energy of a particle executing simple harmonic motion is also periodic, with period \(T / 2\), being zero at the mean position and maximum at the extreme displacements.
It follows from Eqs. (14.15) and (14.17) that the total energy, \(E\), of the system, is,
\(
\begin{aligned}
& E=U+K \\
& =\frac{1}{2} k A^2 \cos ^2(\omega t+\phi)+\frac{1}{2} k A^2 \sin ^2(\omega t+\phi) \\
& =\frac{1}{2} k A^2\left[\cos ^2(\omega t+\phi)+\sin ^2(\omega t+\phi)\right]
\end{aligned}
\)
Using the familiar trigonometric identity, the value of the expression in the brackets is unity. Thus,
\(
E=\frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 \dots(14.18)
\)
The total mechanical energy of a harmonic oscillator is thus independent of time as expected for motion under any conservative force. The time and displacement dependence of the potential and kinetic energies of a linear simple harmonic oscillator are shown in Fig. 14.16.

Observe that both kinetic energy and potential energy in SHM are seen to be always positive in Fig. 14.16. Kinetic energy can, of course, be never negative, since it is proportional to the square of speed. Potential energy is positive by choice of the undermined constant in potential energy. Both kinetic energy and potential energy peak twice during each period of SHM. For \(x=0\), the energy is kinetic; at the extremes \(x= \pm A\), it is all potential energy. In the course of motion between these limits, kinetic energy increases at the expense of potential energy or vice-versa.

Example 1: A particle of mass \(0.50 \mathrm{~kg}\) executes a simple harmonic motion under a force \(F=-\left(50 \mathrm{~N} \mathrm{~m}^{-1}\right) x\). If it crosses the centre of oscillation with a speed of \(10 \mathrm{~m} \mathrm{~s}^{-1}\), find the amplitude of the motion.

Solution:

The kinetic energy of the particle when it is at the centre of oscillation is \(E=\frac{1}{2} m v^2\)
\(
\begin{aligned}
& =\frac{1}{2}(0 \cdot 50 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2 \\
& =25 \mathrm{~J} .
\end{aligned}
\)
The potential energy is zero here. At the maximum displacement \(x=A\), the speed is zero and hence the kinetic energy is zero. The potential energy here is \(\frac{1}{2} k A^2\). As there is no loss of energy,
\(
\frac{1}{2} k A^2=25 \mathrm{~J} \dots(i)
\)
The force on the particle is given by
\(
F=-\left(50 \mathrm{~N} \mathrm{~m}^{-1}\right) x \text {. }
\)
Thus, the spring constant is \(k=50 \mathrm{~N} \mathrm{~m}^{-1}\).
Equation (i) gives
\(
\frac{1}{2}\left(50 \mathrm{~N} \mathrm{~m}^{-1}\right) A^2=25 \mathrm{~J}
\)
or,
\(
A=1 \mathrm{~m} .
\)

Example 2: A particle of mass \(200 \mathrm{~g}\) executes a simple harmonic motion. The restoring force is provided by a spring of spring constant \(80 \mathrm{~N} \mathrm{~m}^{-1}\). Find the time period.

Solution:

The time period is
\(
T=2 \pi \sqrt{\frac{m}{k}}
\)
\(
\begin{aligned}
& =2 \pi \sqrt{\frac{200 \times 10^{-3} \mathrm{~kg}}{80 \mathrm{Nm}^{-1}}} \\
& =2 \pi \times 0^{.} 05 \mathrm{~s}=0.31 \mathrm{~s} .
\end{aligned}
\)

Example 3: A block whose mass is \(1 \mathrm{~kg}\) 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 \(5 \mathrm{~cm}\) away from the mean position.

Solution:

The block executes SHM, and its angular frequency, as given by Eq. (14.14b), is
\(
\begin{aligned}
\omega & =\sqrt{\frac{k}{m}} \\
& =\sqrt{\frac{50 \mathrm{~N} \mathrm{~m}^{-1}}{1 \mathrm{~kg}}} \\
& =7.07 \mathrm{rad} \mathrm{s}^{-1}
\end{aligned}
\)
Its displacement at any time t is then given by,
\(
x(t)=0.1 \cos (7.07 t)
\)
Therefore, when the particle is \(5 \mathrm{~cm}\) 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
\(
\begin{aligned}
& =0.1 \times 7.07 \times 0.866 \mathrm{~m} \mathrm{~s}^{-1} \\
& =0.61 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
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 \(x=5 \mathrm{~cm}\),
\(
\begin{aligned}
& =\text { K.E. }+ \text { P.E. } \\
& =0.25 \mathrm{~J}
\end{aligned}
\)
we also know that at maximum displacement, \(\mathrm{K}\). \(\mathrm{E}\). 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 the same as the sum of the two energies at a displacement of \(5 \mathrm{~cm}\). This is in conformity with the principle of conservation of energy.

Example 4: A particle executes simple harmonic motion of amplitude A along the X-axis. At \(t=0\), the position of the particle is \(x=A / 2\) and it moves along the positive \(x\)-direction. Find the phase constant \(\delta\) if the equation is written as \(x=A \sin (\omega t+\delta)\).

Solution:

We have \(x=A \sin (\omega t+\delta)\). At \(t=0, x=A / 2\).
Thus, \(\quad A / 2=A \sin \delta\)
or, \(\quad \sin \delta=1 / 2\)
or, \(\delta=\pi / 6\) or \(5 \pi / 6\).
The velocity is \(v=\frac{d x}{d t}=A \omega \cos (\omega t+\delta)\).
At \(t=0, v=A \omega \cos \delta\).
Now, \(\quad \cos \frac{\pi}{6}=\frac{\sqrt{ } 3}{2}\) and \(\cos \frac{5 \pi}{6}=-\frac{\sqrt{ } 3}{2}\).
As \(v\) is positive at \(t=0\), \(\delta\) must be equal to \(\pi / 6\).

Example 5: A particle of mass \(40 \mathrm{~g}\) executes a simple harmonic motion of amplitude \(2.0 \mathrm{~cm}\). If the time period is \(0.20 \mathrm{~s}\), find the total mechanical energy of the system.

Solution:

The total mechanical energy of the system is
\(
\begin{aligned}
E & =\frac{1}{2} m \omega^2 A^2 \\
& =\frac{1}{2} m\left(\frac{2 \pi}{T}\right)^2 A^2=\frac{2 \pi^2 m A^2}{T^2} \\
& =\frac{2 \pi^2\left(40 \times 10^{-3} \mathrm{~kg}\right)\left(2 \cdot 0 \times 10^{-2} \mathrm{~m}\right)^2}{(0 \cdot 20 \mathrm{~s})^2} \\
& =7 \cdot 9 \times 10^{-3} \mathrm{~J} .
\end{aligned}
\)