A string vibrates according to the equation `y=5"sin" (pix)/(3) cos 40 pi t`
where, x and y are in centimeters and t is in seconds.
(a) What is the speed of the component wave?
(b) What is the distance between the adjacent nodes?
(c) What is the velocity of the 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 `