Question

A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m per second and 90 m per second square respectively velocity of the wave is 20 m per second find the waveform

Answer 1

Equation of wave:

y=Asin(kx−ωt+ϕ)=Asin(2πx/λ−2πνt+ϕ)

Differentiate above eq. w.r.t time and get particle velocity,

dy/dt=−Aωcos(kx−ωt+ϕ)

Again differentiate above eq. w.r.t time to get acceleration,

d2y/dt2=−Aω2sin(kx−ωt+ϕ)

Given: ωA=3m/s  and ω2A=90m/s2

ω=30rad/s, ω=30rad/s  

The wave velocity v=νλ=ω/k=20m/s gives k=ω/v=3/2m−1. The wave travelling towards the +x direction is,

y=0.1sin(3x/2−30t+ϕ),

and that travelling towards the −x direction is,

y=0.1sin(3x/2+30t+ϕ)

Answer 2

Answer:

y=0.1sin(30t±1.5x)

Explanation:

y=Asin(kx±ωt)

v$_{max}$=Aω=3m$s^{-1}$                          _1

a$_{max}$=Aω^2=90m$s^{-2}$                   _2

v=ω/k=20m$s^{-1}$                            _3

from 1 & 2,

   A=(v$_{max}$)/a$_{max}$=9/90=0.1m

   ω=30rad$s^{-1}$

now in 3,

   k=30/20=1.5rad$m^{-1}$

Therefore, y=0.1sin(30t±1.5x)

TAKE CARE.

HOPE THIS WILL HELP YOU.