2.1 Transverse harmonic waves  (Page 3/6)

$$\sin \theta =\sin \left(\pi -\theta \right)$$

$$\sin \frac{\pi }{3}=\sin \left(\pi -\frac{\pi }{3}\right)=\sin \left(\frac{2\pi }{3}\right)=\frac{1}{2}$$

We can determine the slope of the waveform by partially differentiating y-function with respect to "x" (considering "t" constant) :

$$\Rightarrow \frac{\partial y}{\partial t}=\frac{\partial }{\partial t}A\sin \left(kx-\omega t+\phi \right)=kA\cos \left(kx-wt+\phi \right)$$

$$\text{At}x=\mathrm{0,}t=\mathrm{0,}\phi =\pi /3$$

$$\Rightarrow \text{Slope}=kA\cos \phi =kA\cos \frac{\pi }{3}=kAX\frac{\sqrt{3}}{2}=\text{a positive number}$$

$$\text{At}x=\mathrm{0,}t=\mathrm{0,}\phi =2\pi /3$$

$$\Rightarrow \text{Slope}=kA\cos \phi =kA\cos \frac{2\pi }{3}=-kAX\frac{\sqrt{3}}{2}=\text{a negative number}$$

The smaller of the two (π/3) in first quadrant indicates that wave form has positive slope and is increasing as we move along x-axis. The greater of the two (2π/3) similarly indicates that wave form has negative slope and is decreasing as we move along x-axis. The two initial wave forms corresponding to two initial phase angles in first and second quadrants are shown in the figure below.

We can interpret initial phase angle in the third and fourth quadrants in the same fashion. The sine values of angles in third and fourth quadrants are negative. There is a pair of two angles for which sine has equal negative values. The angle in third quarter like 4π/3 indicates that wave form has negative slope and is further decreasing (more negative) as we move along x-axis. On the other hand, corresponding angle in fourth quadrant for which magnitude is same is 5π/3.

$$\sin \frac{4\pi }{3}=\sin \frac{5\pi }{3}=-\frac{1}{2}$$

For initial phase angle of 5π/3 in fourth quadrant, the wave form has positive slope and is increasing (less negative) as we move along x-axis. The two initial wave forms corresponding to two initial phase angles in third and fourth quadrants are shown in the figure below.

We can also denote initial phase angles in third and fourth quadrants (angles greater than “π”) as negative angles, measured clockwise from the reference direction. The equivalent negative angles for the example here are :

$$-\left(2\pi -\frac{4\pi }{3}\right)=\frac{4\pi }{3}-2\pi =-\frac{2\pi }{3}$$

and

$$-\left(2\pi -\frac{5\pi }{3}\right)=\frac{5\pi }{3}-2\pi =-\frac{\pi }{3}$$

Particle velocity and acceleration

Particle velocity at a given position x=x is obtained by differentiating wave function with respect to time “t”. We need to differentiate equation by treating “x” as constant. The partial differentiation yields particle velocity as :

$$v_p=\frac{\partial }{\partial t}y\left(x,t\right)=\frac{\partial }{\partial t}A\sin \left(kx-\omega t\right)=-\omega A\cos \left(kx-\omega t\right)$$

We can use the property of cosine function to find the maximum velocity. We obtain maximum speed when cosine function evaluates to “-1” :

$$\Rightarrow v_{p\text{max}}=\omega A$$

The acceleration of the particle is obtained by differentiating expression of velocity partially with respect to time :

$$\Rightarrow a_p=\frac{\partial }{\partial t}{v}_p=\frac{\partial }{\partial t}\{-\omega A\cos \left(kx-\omega t\right)\}=-{\omega }^{2}A\sin \left(kx-\omega t\right)=-{\omega }^{2}y$$

Again the maximum value of the acceleration can be obtained using property of sine function :

$$\Rightarrow a_{p\text{max}}={\omega }^{2}A$$

Relation between particle velocity and wave (phase) speed

We have seen that particle velocity at position “x” and time “t” is obtained by differentiating wave equation with respect to “t”, while keeping time “x” constant :

$$v_p=\frac{\partial }{\partial t}y\left(x,t\right)=-\omega A\cos \left(kx-\omega t\right)$$

On the other hand, differentiating wave equation with respect to “x”, while keeping “t”, we have :

$$\frac{\partial }{\partial x}y\left(x,t\right)=kA\cos \left(kx-\omega t\right)$$

The partial differentiation gives the slope of wave form. Knowing that speed of the wave is equal to ratio of angular frequency and wave number, we divide first equation by second :