For a certain transverse standing wave on a long string , an antinode is formed at ` x = 0` and next to it , a node is formed at `x = 0.10 m` , the displacement `y(t)`of the string particle at `x = 0` is shown in Fig.7.97.

Options

A. Transverse displacement of the particle at `x = 0.05 m` and `t = 0.05 s is - 2 sqrt(2) cm`
B. Transverse displacement of the particle at `x = 0.04 m` and `t = 0.025 s is - 2 sqrt(2) cm`
C. Speed of the travelling waves that interface to produce this standing wave is `2 m//s`
D. The transverse velocity of the string particle at `x = 1//15 m` and `t = 0.1 s is 20 pi cm//s`

Answer

A, C, D

Explanation

`(lambda)/(4) = 0.1 rArr lambda = 0.4`
From graph ` rArr T = 0.2 s` and amplitude of standing wave is ` 2 A = 4 cm`.
Equation of the standing wave
`y ( x, t) = - 2 A cos (( 2pi)/(0.4) x) sin (( 2 pi)/(0.2) t) cm`
`Y( x = 0.05 , t = 0.05) = - 2 sqrt(2) cm`
`Y(x = 0.04 , t = 0.05) = - 2 sqrt(2) cos 36^(@)`
Speed ` = (lambda)/( T) = 2 m//s`
`V_(y) = ( dy)/( dt) = - 2 A xx ( 2pi)/(0.2) cos (( 2 pi x)/( 0.4)) cos ((2 pi t)/(0.2))`
`V_(y)(x = (1)/(15) m , t = 0.1) = 20 pi cm//s`