2.1 Transverse harmonic waves  (Page 2/6)

It should be clearly understood that repetition of phase (displacement) is not identified by the magnitude of displacement alone. In the figure shown below, the points “A”, “B” and ”C” have same y-displacements. However, the phase of the waveform at “B” is not same as that at “A”. Note that sense of vibration at “A” and “B” are different (As a matter of fact, the particle at "A" is moving down whereas particle at "B" is moving upward. We shall discuss sense of vibrations at these points later in the module). Since phase is identified by the y-displacement and sense of vibration at a particular position, we need to compare the sense along with the magnitude of y-displacement as well to find the wavelength on a waveform. The wavelength, therefore, is equal to linear distance between positions of points “A” and “C”.

$$AC=\lambda$$

Mathematically, let us consider y-displacement at a position specified by x=x. At time t = 0, the y-displacement is given by :

$$y\left(x=x,t=0\right)=A\sin kx$$

According to definition of wavelength, the y-displacement is same at a further distance “λ”. It means that :

$$\Rightarrow y\left(x=x,t=0\right)=y\left(x=x+\lambda ,t=0\right)$$

$$\Rightarrow A\sin kx=A\sin k\left(x+\lambda \right)=A\sin \left(kx+k\lambda \right)$$

The equality given above is valid when :

$$\Rightarrow k\lambda =2\pi$$

$$\Rightarrow k=\frac{2\pi }{\lambda }$$

Speed of transverse harmonic wave

We can determine speed of the wave by noting that wave travels a linear distance “λ” in one period (T). Thus, speed of wave is given by :

$$v=\frac{\lambda }{T}=\nu \lambda =\frac{\frac{2\pi }{k}}{\frac{2\pi }{\omega }}=\frac{\omega }{k}$$

It is intuitive to use the nature of phase to determine wave speed. We know that phase corresponding to a particular disturbance is a constant :

$$kx-\omega t=\text{constant}$$

Speed of wave is equal to time rate at which disturbance move. Hence, differentiating with respect to time,

$$\Rightarrow k\frac{dx}{dt}-\omega =0$$

Rearranging, we have :

$$\Rightarrow v=\frac{dx}{dt}=\frac{\omega }{k}$$

Here, “ω” is a SHM attribute and “k” is a wave attribute. Clearly, wave speed is determined by combination of SHM and wave attributes.

Initial phase

At x=0 and t =0, the sine function evaluates to zero and as such y-displacement is zero. However, a waveform can be such that y-displacement is not zero at x=0 and t=0. In such case, we need to account for the displacement by introducing an angle like :

$$y\left(x,t\right)=A\sin \left(kx-\omega t+\phi \right)$$

where “φ” is initial phase. At x=0 and t=0,

$$y\left(\mathrm{0,}0\right)=A\sin \left(\phi \right)$$

The measurement of angle determines following two aspects of wave form at x=0, t=0 : (i) whether the displacement is positive or negative and (ii) whether wave form has positive or negative slope.

For a harmonic wave represented by sine function, there are two values of initial phase angle for which displacement at reference origin (x=0,t=0) is positive and has equal magnitude. We know that the sine values of angles in first and second quadrants are positive. A pair of initial phase angles, say φ = π/3 and 2π/3, correspond to equal positive sine values as :