Solved Example Problems for Progressive Waves (or) Travelling Waves

EXAMPLE 11.11

Sketch y = x âˆ’a for different values of a.

Solution

This implies, when increasing the value of a, the line shifts towards right side. For a = vt, y = x âˆ’ vt satisfies the differential equation. Though this function satisfies the differential equation, it is not finite for all values of x and t. Hence, it does not represent a wave.

EXAMPLE 11.12

How does the wave y = sin(x âˆ’ a) for a = 0, a = Ï€/4, a = Ï€/2, a = 3Ï€/2, a = Ï€ look like?. Sketch this wave.

Solution

From the above picture we observe that y = sin (x−a) for a = 0, a = Ï€/4, a = Ï€/2, a = 3Ï€/2 and a = Ï€, the function y = sin (x−a) shifts towards right. Further, we can take a = vt and v = Ï€/4 ,  and sketching for different times t = 0s, t = 1 s, t = 2s etc., we once again observe that y = sin(x−vt) moves towards the right. Hence, y = sin(x−vt) is a travelling (or progressive) wave moving towards the right. If y = sin(x+vt) then the travelling (or progressive) wave moves towards the left. Thus, any arbitrary function of type y = f(x−vt) characterising the wave must move towards right and similarly, any arbitrary function of type y = f(x+vt) characterizing the wave must move towards left.

EXAMPLE 11.13

Check the dimensional of the wave y = sin(x−vt). If it is dimensionally wrong, write the above equation in the correct form.

Solution

Dimensionally it is not correct. we know that y = sin(x−vt) must be a dimensionless quantity but x−vt has dimension. The correct equation is y = sin (k x−ωt), where k and Ï‰ have the dimensions of inverse of length and inverse of time respectively. The sine functions and cosine functions are periodic functions with period 2Ï€. Therefore, the correct expression is y = sin ([2Ï€/λ]x, [2Ï€/T]t) where λ and T are wavelength and time period, respectively.

In general, y(x,t)=A sin(k x−ωt).