Consider a string fixed at one end . A travelling wave given by the wave equation ` y = A sin ( omega t - k x)` is incident on it . ltbr gt Show that at the fixed end of a string the waves suffers a phase change of `pi`, i.e., as it travels back as if the wave is inverted.

Explanation

Wave equation of incident wave is
`y_(1) A sin ( omega t - kx) ( "positive x - direction")`
When the ave strikes the fixed end it must be reflected. The wave will now be travelling in the negative x - direction , and so its equation is
` y_(2) A sin ( omega t + kx + phi ) ( "negative x - direction")`
The phase constant has been added to account for any phase change after reflection . Let us take ` x = 0 ` at the fixed end . The behaviour of a wave at particular positions is governed by appropriate boundary conditions . For fixed ends , the boundary condition is that the end point is a node .
The resultant motion due to incident and reflected wave is ` y = y_(1) + y_(2)`
`y = A [ sin ( omega t - kx) + sin ( omega t + kx + phi)]` (i)
The boundary condition that must be satisfied by the resultant wave on string is ` y = 0 at x = 0 `
On substituting these values in Eq. (i) , we obtain
`sin omega t + sin ( omega t + phi ) = 0 `
`sin omega t = - sin ( omega t + phi)` (ii)
Equation (ii) must be satisfied at all times . For the sake of convenience , we take ` omega t = pi //2`.
`sin ((pi)/(2) + phi)= -1`
`sin theta = -1 ` implies `theta` can have any of the following values.
`3 pi//2 , 7 pi //2 , 11 pi//2 ,....`
Therefore `phi` can have any of the values `pi , 3 pi , 5 pi` , etc. All these values are physically possible and distinguishable . We choose the simplest one , so `phi = pi`.