A stationary wave is given by
`y = 5 sin (pi x)/( 3) cos 40 pi t`
where `x and y` are in `cm and t` is in second.
a. What are the amplitude and velocity of the component wave whose superposition can give rise to this vibration ?
b. What is the distance between the nodes ?
c. What is the velocity of a particle of the string at the position ` x = 1.5 cm when t = 9//8 s`?

Answer

A, B, C

Explanation

(a) `v= omega/k = (40pi)/(pi//3)`
` = 120 cm//s`
(b) Distance between adjacent nodes
` = lambda/2 = pi/k = (pi/(pi//3))`
= 3 cm
(c) v_P = (dely)/(delt) = (-200pi) sin (pix)/3 sin 40pit`
Now, substitute the given value of x and t.