14.5 Phase

14.5 Phase

The angle $\theta = \omega t$ which specifies the displacements $x$ as well as the direction of motion of the point oscillating S.H.M. is called phase.

Phase is the quantity which shows the state of motion of an oscillator. In circular motion, the displacement of the projection of the body moving in a circle, executing S.H.M. on the diameter of the circle, is given by: $x = x_{o}\cos\theta$

or $x = x_{o}\cos(\omega t + 180^{\circ}) \quad \ldots (14.17)$

Where $x$ is instantaneous displacement, $x_{o}$ is maximum displacement; $\omega$ is the angular velocity its graphical representation is shown in Fig. 14.8 (a). The general way of showing this equation is:

The time varying quantity $(\omega t + \phi)$, is called the phase of the motion. It describes the state of motion at a given time. The constant $\phi$ is called the phase constant (or phase angle). The value of the $\phi$ depends on displacement and velocity of the particle at $t = 0$.

The quantity $\phi$ represents the phase difference between the states of motion of two oscillators. Let us explain, with the help of graph plotted between $x$ and $t$.

If $\phi = 0^{\circ}$, then Eq. (14.18) becomes:

$x = x_{o}\cos\omega t$

Putting $t = 0, T/4, T/2, 3T/4, \dots$, we obtain a graph, as shown in Fig. 14.8 (a). This graph shown that;

at $t = 0, T/2$ and $T$ (corresponding to $\theta = 0, \pi$ and $2\pi$), the point is at the extreme positions. At $t = T/4$ and $3T/4$ (corresponding to $\theta = \pi/2$ and $3\pi/2$), the point is at mean position.

(Fig. 14.8 (a): Graph of $x = x_{o} \cos(\omega t + \phi)$ for $\phi = 0^{\circ}$)

If $\phi = 90^{\circ}$, then Eq. (14.18) becomes:

$x = x_{o}\cos(\omega t + 90^{\circ})$

Putting $t = 0, T/4, T/2, 3T/4, T, \dots$, we obtain graph, as shown in Fig. 14.8 (b). This graph shows that;

at $t = 0, T/2$ and $T$ (corresponding to $\theta = 0, \pi$ and $2\pi$), the point is at mean positions.

At $t = T/4$ and $3T/4$ (corresponding to $\theta = \pi/2$ and $3\pi/2$), the point is at extreme position.

(Fig. 14.8 (b): Graph of $x = x_{o} \cos(\omega t + \phi)$ for $\phi = 90^{\circ}$)

If $\phi = 180^{\circ}$, then Eq. (14.18) becomes:

$x = x_{o}\cos(\omega t + 180^{\circ})$

Putting $t = 0, T/4, T/2, 3T/4, T, \dots$, we obtain a graph, as shown in Fig. 14.8 (c). It can be noted that:

When the phase difference between two oscillating systems is $180^{\circ}$, they are said to be oscillating out of phase.

When phase difference between oscillating systems is $0^{\circ}$ or $360^{\circ}$, they are said to be oscillating in phase.

(Fig. 14.8 (c): Graph of $x = x_{o} \cos(\omega t + \phi)$ for $\phi = 180^{\circ}$)

Example 14.4

A $0.5\,\mathrm{kg}$ mass is attached to a spring with a spring constant of $200\,\mathrm{N\,m^{-1}}$. The mass is displaced by $0.2\,\mathrm{m}$ from its equilibrium position and released from rest. Assuming simple harmonic motion, find:

(a) Displacement at $t = 0.5\,\mathrm{s}$

(b) Velocity at $t = 0.5\,\mathrm{s}$

(c) Acceleration at $t = 0.5\,\mathrm{s}$

Solution

Mass $m = 0.5\,\mathrm{kg}$

Spring constant $k = 200\,\mathrm{N\,m^{-1}}$

Amplitude $x_{o} = 0.2\,\mathrm{m}$

Angular frequency $\omega = ?$

$\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{200\,\mathrm{N\,m^{-1}}}{0.5\,\mathrm{kg}}} = 20\,\mathrm{rad\,s^{-1}}$

(a) Displacement at $t = 0.5\,\mathrm{s}$

$x(t) = x_{o}\cos(\omega t)$

$= 0.2 \cos(20 \times 0.5)$

$= 0.2 \cos(10)$

$= 0.2 \times 0.839$

$= 0.168\,\mathrm{m}$

(b) Velocity at $t = 0.5\,\mathrm{s}$

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

$= 0.2 \times 20 \sin(20 \times 0.5)$

$= 4 \sin(10)$

$v(t) = 4 \times 0.544 = 2.176\,\mathrm{m\,s^{-1}}$

(c) Acceleration at $t = 0.5\,\mathrm{s}$

$a(t) = -x_{o}\omega^{2}\cos(\omega t)$

$= -0.2 \times (20)^{2} \cos(20 \times 0.5)$

$= -80 \cos(10) = -80(0.8391)$

$a(t) = -67.12\,\mathrm{m\,s^{-2}}$