# 1.3 Linear shm  (Page 4/4)

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$⇒\text{Period}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }=\frac{T}{2}$

As time period of variation is half, the frequency of “U” is twice that of displacement. For this reason, potential energy – time plot is denser than that of displacement – time plot.

## Mechanical energy

The basic requirement of SHM is that mechanical energy of the system is conserved. At any point or at any time of instant, the sum of potential and kinetic energy of the system in SHM is constant. This is substantiated by evaluating sum of two energies :

$E=K+U$

Using expressions involving displacement, we have :

$⇒E=\frac{1}{2}m{\omega }^{2}\left({A}^{2}-{x}^{2}\right)+\frac{1}{2}m{\omega }^{2}{x}^{2}=\frac{1}{2}m{\omega }^{2}{A}^{2}$

The plots of kinetic, potential and mechanical energy with respect to displacement are drawn in the figure. Note that the sum of kinetic and potential energy is always a constant, which is equal to the mechanical energy of the particle in SHM.

We can also obtain expression of mechanical energy, using time dependent expressions of kinetic and potential energy as :

$⇒E=\frac{1}{2}m{\omega }^{2}{A}^{2}{\mathrm{cos}}^{2}\left(\omega t+\phi \right)+\frac{1}{2}m{\omega }^{2}{A}^{2}{\mathrm{sin}}^{2}\left(\omega t+\phi \right)$

$⇒E=\frac{1}{2}m{\omega }^{2}{A}^{2}\left\{\mathrm{cos}{}^{2}\left(\omega t+\phi \right)+\mathrm{sin}{}^{2}\left(\omega t+\phi \right)\right\}=\frac{1}{2}m{\omega }^{2}{A}^{2}$

The mechanical energy – time plot is shown in the figure. We observe following important points about variation of energy with respect to time :

• Mechanical energy – time plot is a straight line parallel to time axis. This signifies that mechanical energy of particle in SHM is conserved.
• There is transformation of energy between kinetic and potential energy during SHM.
• At any instant, the sum of kinetic and potential energy is equal to $\frac{1}{2}m{\omega }^{2}{A}^{2}$ or $\frac{1}{2}k{A}^{2}$ , which is equal to maximum values of either kinetic or potential energy.

## Example

Problem 1: The potential energy of an oscillating particle of mass “m” along straight line is given as :

$U\left(x\right)=a+b{\left(x-c\right)}^{2}$

The mechanical energy of the oscillating particle is “E”.

• Determine whether oscillation is SHM?
• If oscillation is SHM, then find amplitude and maximum kinetic energy.

Solution : If the motion is SHM, then restoring force is a conservative force. The potential energy is, then, defined such that :

$dU=-Fdx$

$⇒F=-\frac{dU}{dx}=-2b\left(x-c\right)$

In order to find the center of oscillation, we put F = 0.

$F=-2b\left(x-c\right)=0\phantom{\rule{1em}{0ex}}⇒x-c=0\phantom{\rule{1em}{0ex}}⇒x=c$

This means that particle is oscillating about point x = c. The displacement of the particle in that case is “x-c” – not “x”. This, in turn, means that force is proportional to negative of displacement, “x-c”. Hence, particle is executing SHM.

Alternatively, put y = x-c :

$F=-2by$

This means that particle is executing SHM about y = 0. This means x-c = 0, which in turn, means that particle is executing SHM about x = c.

The mechanical energy is related to amplitude by the relation :

$E=\frac{1}{2}m{\omega }^{2}{A}^{2}$

$⇒A=\sqrt{\left(\frac{2E}{m{\omega }^{2}}\right)}$

Now, $m{\omega }^{2}=k=2b$ . Hence,

$⇒A=\sqrt{\left(\frac{2E}{2b}\right)}=\sqrt{\left(\frac{E}{b}\right)}$

The potential energy is minimum at the center of oscillation i.e. when x = c. Putting this value in the expression of potential energy, we have :

$⇒{U}_{\text{min}}=a+b{\left(c-c\right)}^{2}=a$

It is important to note that minimum value of potential energy need not be zero. Now, kinetic energy is maximum, when potential energy is minimum. Hence,

${K}_{\text{max}}=E-{U}_{\text{min}}=E-a$

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