Question

Show that the equation $y = a\sin (\omega t - kx)$ satisfies the equation$\dfrac{{{\partial ^2}y}}{{\partial {t^2}}} = {v^2}\dfrac{{{\partial ^2}y}}{{\partial {x^2}}}$. Find the speed of the wave and the direction in which it’s travelling.

Answer 1

Hint: Wave: It is a disturbance traveling through a medium, transporting the energy from one location to another location. Waves transport the energy without transporting the matter. Change in the direction of a wave is known as reflection. Waves are shown spectrum like radio waves, gamma waves, visible light etc.

Complete step-by-step solution:
Given,
Equation,
$y = a\sin (\omega t - kx)$ …(1)
$\dfrac{{{\partial ^2}y}}{{\partial {t^2}}} = {v^2}\dfrac{{{\partial ^2}y}}{{\partial {x^2}}}$ …(2)
Differentiate the equation 1 w.r.t. time’t’
$ \Rightarrow y = a\sin (\omega t - kx)$
$ \Rightarrow \dfrac{{\partial y}}{{\partial t}} = a\omega \cos (\omega t - kx)$
Again differentiate w.r.t. ‘t’
$ \Rightarrow \dfrac{{{\partial ^2}y}}{{\partial {t^2}}} = - a{\omega ^2}\sin (\omega t - kx)$ …(3)
Now differentiate the equation 1 w.r.t. ‘x’
$ \Rightarrow y = a\sin (\omega t - kx)$
$ \Rightarrow \dfrac{{\partial y}}{{\partial x}} = - ax\cos (\omega t - kx)$
Differentiate again w.r.t. ‘x’
$ \Rightarrow \dfrac{{{\partial ^2}y}}{{\partial {x^2}}} = - a{x^2}\sin (\omega t - kx)$ …(4)
Put the value in equation 2 from equation 3 and 4
$ \Rightarrow \dfrac{{{\partial ^2}y}}{{\partial {t^2}}} = {v^2}\dfrac{{{\partial ^2}y}}{{\partial {x^2}}}$
$ \Rightarrow - a{\omega ^2}\sin (\omega t - kx) = {v^2}( - a{x^2}\sin (\omega t - kx))$
$ \Rightarrow {\omega ^2} = {v^2}{x^2}$
\[ \Rightarrow {v^2} = {\left( {\dfrac{\omega }{x}} \right)^2}\]
\[ \Rightarrow v = \left( {\dfrac{\omega }{x}} \right)\]
The Waves travel in a positive direction.

Note: Speed of waves depends on the medium in which the wave travels. In denser medium waves travel slow as compared to lesser denser mediums. The velocity of the wave is equal to the product of wavelength and frequency (number of vibrations per second). And it is independent of intensity.