1.1 Simple harmonic motion  (Page 2/4)

Time period

The time period is the time taken to complete one cycle of motion. In our consideration in which we have started observation from positive extreme, this is equal to time taken from start t =0 s to the time when the particle returns to the positive extreme position again.

$$\text{At}t=\mathrm{0,}\omega t=\mathrm{0,}\cos \omega t=\cos 0^0=1$$

$$\Rightarrow x=A$$

$$\text{At}t=\frac{2\pi }{\omega },\omega t=\omega X\frac{2\pi }{\omega }=2\pi ,\cos \omega t=\cos 2\pi =\mathrm{1,}$$

$$\Rightarrow x=A$$

Thus, we see that particle takes a time “2π/ω” to return to the extreme position from which motion started. Hence, time period of SHM is :

$$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi }{2\pi \nu }=\frac{1}{\nu }$$

Shm and uniform circular motion

The expression for the linear displacement in “x” involves angular frequency :

$$x=A\cos \omega t$$

Clearly, we need to understand the meaning of angular frequency in the context of linear “to and fro” motion. We can understand the connection or the meaning of this angular quantity knowing that SHM can be interpreted in terms of uniform circular motion. Consider SHM of a particle along x-axis, while another particle moves along a circle at a uniform angular speed, "ω", in anticlockwise direction as shown in the figure. Let both particles begin moving from point “P” at t=0.

After time “t”, the particle executing uniform circular motion (UCM), covers an angular displacement “ωt” and reaches a point “Q” as shown in the figure. The projection of line joining origin, "O" and “Q” on x – axis is :

$$OR=x=A\cos \omega t$$

Now compare this expression with the expression of displacement of particle executing SHM. Clearly, both are same. It means projection of the position of the particle executing UCM is equal to the displacement of the particle executing SHM from the origin. This is the connection between two motions. Also, it is obvious that angular frequency “ω” is equal to the uniform angular speed, “ω" of the particle executing UCM .

In case, if this analogical interpretation does not help to interpret angular frequency, then we can simply think that angular frequency is product of “2π” and frequency “ν”.

$$\omega =2\pi \nu$$

Phase constant

We used a cosine function to represent displacement of the particle in SHM. This function represents displacement for the case when we start observing motion of the particle at positive extreme. At t = 0,

$$x=A\cos \omega t=A\cos 0^0=A$$

What if we want to observe motion from the position when the particle is at mean position i.e. at “O”. We know that sine of zero is zero. Knowing the nature of sine curve, we can ,intuitively, say that sine function would fit in the requirement in this case and displacement is given as :

$$x=A\sin \omega t$$

However, if we want to stick with the cosine function, then there is a way around. We know that :

$$\cos \theta =\sin \left(\theta \pm \frac{\pi }{2}\right)$$

Keeping this in mind, we represent the displacement as :

$$\Rightarrow x=A\cos \left(\omega t\pm \frac{\pi }{2}\right)$$

Let us check out the position of the particle at t = 0,

$$\Rightarrow x=A\cos \left(\pm \frac{\pi }{2}\right)=0$$

Clearly, cosine function represents displacement with this modification even in case when we start observing motion of the particle at mean position. In other words, cosine function, as modified, is equivalent to sine function.