Two sinusoidal waves combining in a medium are described by the equations
`y_1 = (3.0 cm) sin pi (x+ 0.60t)`
and `y_2 = (3.0 cm) sin pi (x-0.06 t)`
where, x is in centimetres and t is in seconds. Determine the maximum displacement of the medium at
(a)x=0.250 cm,
(b)x=0.500 cm and
(c) x=1.50 cm.
(d) Find the three smallest values of x corresponding to antinodes.

Explanation

The answers for maximum transverse position must be between ` 6 cm and 0`. The anitodes are seperated by half a wavelength , which we except to be a couple of centimetres .
According to the waves in interference model , we write the function ` y_(1) + y_(2)` and start evaluating things.
We get ` y = y_(1) + y_(2) = ( 6.0 cm) sin ( pi x) cos ( 0.60 pi t)`
Since ` cos (0) = 1` , we can find the minimum value of `y` by setting ` t = 0` .
`y_(max) (x) = y_(1) + y_(2) = ( 6.0 cm) sin (pi x)`
(a) At `x = 0.250 cm , y_(max) = ( 6.0 cm) sin ( 0.250 pi) = 4.24 cm`
(b) At ` x= 0.50 cm , y_(max) = ( 6.0 cm) sin ( 0.500 pi ) = 6.00 cm`
( c) At ` x= 1.50 cm , y _(max) = | ( 6.0 cm) sin (1.50 pi)| = + 6.00 cm`
(d). The antinodes occur when ` x = n lambda //4 for n = 1 , 3 , 5 , .....`
But ` k = 2 pi //lambda = pi , so lambda = 2.00 cm `
and ` x_(1) = lambda//4 = ( 2.00 cm)//4 = 0.500 cm`
`x_(2) = 3 lambda //4 = 3 ( 2.00 cm )//4 = 1.50 cm`
` x_(3) = 5 lambda //4 = 5 ( 2.00 cm)//4 = 2.50 cm`
Two of our answers in (d) can be read from the way the amplitude had its largest possiblr value in parts (b) and ( c ). Note again that an amplitude is defined to be always positive , as the maximum absolute value of wave function .