1.3 Linear shm  (Page 3/4)

$$\Rightarrow K_{\text{min}}=\frac{1}{2}m{\omega }^2\left(A^2-A^2\right)=0$$

By substituting for “x” in the equation of kinetic energy, we get expression of kinetic energy in terms of variable time, “t” as :

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

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

1: The KE function is square of cosine function. It means that KE is always positive.

2: The time period of KE is half that of displacement. We know the trigonometric identity :

$${\cos }^{2}x=\frac{1}{2}\left(1+\cos 2x\right)$$

Applying this trigonometric identity to the square of cosine term in the expression of kinetic energy as :

$$\Rightarrow K=\frac{1}{4}m{\omega }^2{A}^2\{1+\cos 2\left(\omega t+\phi \right)\}=\frac{1}{4}m{\omega }^2{A}^2\{1+\cos \left(2\omega t+2\phi \right)\}$$

Applying rules for finding time period, we know that period of function “kf(x)” is same as that of “f(x)”. Hence, period of “K” is same as that of “1 + cos (2ωt + 2φ)”. Also, we know that period of function “f(x) + a” is same as that of “f(x)”. Hence, period of “K” is same as that of “cos (2ωt + 2φ)”. Now, period of “f(ax±b)” is equal to period of “f(x)” divided by “|a|”. Hence, period of “K” is :

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

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

Potential energy

We recall that potential energy is an attribute of conservative force system. The first question that we need to answer is whether restoring force in SHM is a conservative force? One of the assumptions, which we made in the beginning, is that there is no dissipation of energy in SHM. It follows, then, that restoring force in SHM is a conservative force.

Second important point that we need to address is to determine a reference zero potential energy. We observe that force on the particle in SHM is zero at the center and as such serves to become the zero reference potential energy. Now, potential energy at a position “x” is equal to negative of the work done in taking the particle from reference point to position “x”.

$$U=-W=-\int Fdx=-\int \left(-kx\right)dx=k\int xdx$$

Integrating in the interval, we have :

$$\Rightarrow U=k\underset{0}{\overset{x}{\int }}xdx=k[\frac{x^2}{2}\underset{0}{\overset{x}{]}}=\frac{1}{2}k{x}^2$$

Thus, instantaneous potential energy of oscillating particle is given as :

$$\Rightarrow U=\frac{1}{2}k{x}^2=\frac{1}{2}m{\omega }^2{x}^2$$

The maximum value of PE corresponds to position when speed is zero. At x = A,

$$\Rightarrow U_{\text{max}}=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2$$

The minimum value of PE corresponds to position when speed has maximum value. At x = 0,

$$\Rightarrow U_{\text{min}}=\frac{1}{2}kX{0}^2=0$$

By substituting for “x” in the equation of potential energy, we get expression of kinetic energy in terms of variable time, “t” as :

$$U=\frac{1}{2}k{A}^2\sin {}^2\left(\omega t+\phi \right)=\frac{1}{2}m{\omega }^2{A}^2\sin {}^2\left(\omega t+\phi \right)$$

The potential energy – time plot is shown in the figure. We observe following important points about variation of kinetic energy :

1: The KE function is square of sine function. It means that PE is always positive.

2: The time period of KE is half that of displacement. We have already proved the same in the case of kinetic energy. We can extend the reason in the case of potential energy as well :