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`?

Explanation

Using the relation ` 2 sin C cos D = sin ( C + D) + sin ( C - D)`
`y = 5 sin (pi x)/(3) cos 40 pi t = (5)/(2) sin ( pi x)/(3) cos 40 pi t`
`y = (5)/(2) [ sin (( pi x)/( 3) + 40 pi t) + sin (( pi x)/( 3) - 40 pi t)]`
`= (5)/(2) sin ( 40 pi t + ( pi x)/(3)) - (5)/(2) sin ( 40 pi t - (pi x)/( 3))`
Thus , the given stationary wave is formed by the superposition of the progressive waves
` y_(1) = (5)/(2) sin ( 40 pi t + (pi x)/( 3))` ltbrlt and `y_(2) = (5)/(2) sin ( 40 pi t - ( pi x)/(3) + pi)`
a. Comparing each wave with the standard form of the progressive wave
` y = a sin ( omega t - ( 2 pi)/( lambda) + alpha)`
` a = 5//2 = 2.5 cm`
` omega = 40 pi rArr 2 pi f = 40 rarr f = 20 s^(-1)`
and ` ( 2pi)/( lambda) = (pi)/( 3) or lambda = 6 cm = 0.06 m`
`:. c = f lambda = 20 xx 0.06 = 1.2 m//s`
b. Distance between the nodes ` = lambda //2 = 0.06//2 = 0.03 m`
c. `y = 5 sin ( pi x)/(3) cos 40 pi t`
`v = ( dy)/( dt) = -5 xx 40 pi sin (pi x)/(3) 40 pi t`
`rArr v = - 200 ( pi x)/(3) sin 40 pi t`
`:. At x = 1.5 cm and t = 9//8 s`
d. ` v = - 200 pi ( pi//2) sin 45 pi = 0 `