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 `-2sqrt2` cm
B. Transverse displacement of the particle at x=0.04 m and t=0.025 s is `-2sqrt2` 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 `20pi` cm/s

Explanation

`lambda/4=0.1 rArr lambda=0.4 m`
from graph `rArr` T=0.2 sec and amplitude of standing wave is `2A=4 cm`
Equation of the standing wave
`y(x,t)=-2A cos ((2pi)/(0.4)x)sin((2pi)/(0.2)t)` cm
`y(x=0.05, t=0.05=-2sqrt2` cm
`y(x=0.05, t=0.25=-2sqrt2 cos 36^@`
speed =`lambda/T`=2m/sec
`V_y=(dy)/(dt)=-2Axx(2pi)/0.2cos((2pix)/0.4)cos ((2pit)/(0.2))`
`V_y=(x=1/15m, t=0.1)=20 pi` cm/sec.