14.7 Conservation of Energy in S.H.M

14.7 Conservation of Energy in S.H.M

When a body is executing simple harmonic motion, it possesses both potential energy as well as kinetic energy. Its potential energy is on account of its displacement from mean position and its kinetic energy is due to its velocity. These energies vary during the oscillation, but the total energy at any instant remains constant in the absence of unbalanced frictional forces. In case of mass-spring system shown in Fig. (14.10), when the mass is displaced from the mean position O then there is a restoring force ($F$) whose value is zero at mean position when $x = 0$ and its value is maximum at either extreme position where $x = x_{o}$. Thus, average value of force from the mean position to the extreme position is:

When displacement $= 0$ ; Force $= 0$

When displacement $= x_{o}$ ; Force $= F = kx_{o}$

$F_{\mathrm{av}} = \dfrac{0 + kx_{o}}{2}$

$F_{\mathrm{av}} = \dfrac{1}{2}kx_{o}$

When the spring is stretched to its maximum displacement $x_{o}$, work is done on the spring which is given by

This work done on the mass attached to a spring is stored in terms of potential energy, called elastic potential energy. So, we have the expression for P.E. at any instant $x$ given by

It is clear from Eq. (14.22) that the potential energy of oscillator is zero at $x = 0$ and maximum at $x = \pm x_{o}$ i.e., the extreme position on either side and total energy is entirely elastic potential energy.

After the removal of force, the mass attached to a spring starts its motion with velocity $v$.

As the instantaneous velocity $v$ of the mass executing simple harmonic motion is given by;

As $\omega = \sqrt{\dfrac{k}{m}}$ , so above Eq. becomes:

At mean position $x = 0$, the mass gets maximum velocity i.e.,

Hence, at any instant where the displacement is $x$, the kinetic energy of mass attached to spring is given by

At mean position ($x = 0$), the elastic P.E. is zero but the velocity is maximum, hence, total energy is entirely kinetic energy.

The total energy is the sum of kinetic and potential energy of the system at displacement '$x$', so we may write;

$\mathrm{T.E} = \mathrm{K.E} + \mathrm{P.E}$

Substituting the values of K.E. and P.E. from above Eqs., we have

As, $\omega = \sqrt{\dfrac{k}{m}}$ or $k = m\omega$, so Eq. 14.28 can be written as;

From Eqs. (14.23), (14.27) and (14.28), it is proved that total energy of mass spring system is constant and is directly proportional to the square of the amplitude of oscillation. This statement of conservation of energy is equally valid for all bodies executing S.H.M. When K.E. of mass is maximum, mass passes through the centre of oscillation, the P.E. of mass spring is zero ($x = 0$), conversely when the P.E. of spring is maximum, the mass is at its extreme position on either side, the K.E. of mass is zero ($x = x_{o}$).

The energy oscillates between K.E. and P.E., but their sum remains constant. This can be illustrated by plotting graph of K.E., P.E. and T.E. verses displacement as shown in Fig. (14.11).

(Fig. 14.11: Graphical representation of energy (K.E, P.E. and T.E) against displacement, for S.H.M)

Example 14.5

A body of $0.010\,\mathrm{kg}$ executes simple harmonic motion of period $6\,\mathrm{s}$. one second after it has passed its mean position, it's velocity is $0.06$ meter per second. Find its kinetic energy, potential energy and Total energy.

Solution

$m = 0.010\,\mathrm{kg}$

$T = 6\,\mathrm{s}$

$v = 0.06\,\mathrm{m\,s^{-1}}$ (at $t = 1\,\mathrm{s}$)

Angular frequency ($\omega$)

$\omega = 2\pi/T = 2\pi/6 = \pi/3\,\mathrm{rad\,s^{-1}}$

$\mathrm{K.E.} = (\tfrac{1}{2})mv^{2}$

$= (\tfrac{1}{2}) \times 0.010\,\mathrm{kg} \times (0.06\,\mathrm{m\,s^{-1}})^{2}$

$= 1.8 \times 10^{-5}\,\mathrm{J}$

To find the total energy ($E_{T}$), first, find the amplitude $x_{o}$, using the formula of velocity;

$v = x_{o}\omega\sin(\omega t)$

$0.06\,\mathrm{m\,s^{-1}} = x_{o}\left(\dfrac{\pi}{3}\right)\sin\left(\dfrac{\pi}{3} \times 1\right)$

$0.06\,\mathrm{m\,s^{-1}} = x_{o}\dfrac{\pi}{3}(0.866)$

$0.06\,\mathrm{m\,s^{-1}} = x_{o}(1.047)(0.866)$

$0.06\,\mathrm{m\,s^{-1}} = 0.90x_{o}$

$x_{o} = 0.066\,\mathrm{m}$

Total energy $E_{T}$ of system is given by

$E_{T} = \dfrac{1}{2}m\omega^{2}x_{o}^{2}$

Substituting the values, we have:

$E_{T} = \dfrac{1}{2}0.01\left(\dfrac{\pi}{3}\right)^{2}(0.066)^{2}$

$E_{T} = 2.4 \times 10^{-5}\,\mathrm{J}$

$\mathrm{P.E.} = E_{T} - \mathrm{K.E}$

$= 2.4 \times 10^{-5} - 1.8 \times 10^{-5}$

$\mathrm{P.E.} = 0.6 \times 10^{-5}\,\mathrm{J}$

$\mathrm{P.E.} = \dfrac{1}{2}kx^{2} = \dfrac{1}{2}m\omega^{2}x^{2}$

Note:

(i) P.E. can also be calculated by expression $\mathrm{P.E.} = \dfrac{1}{2}kx^{2} = \dfrac{1}{2}m\omega^{2}x^{2}$ where $x$ being displacement of a particle which can be found out by $x = x_{o}\cos\omega t$.

(ii) The total energy of a system can also be found out by $E_{T} = \mathrm{K.E.} + \mathrm{P.E}$.