Question

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is

y(x,t) = 2.35 mm cos[(7.00rad/m)x + (753 rad/s)t].

Being more practical, you measure the rope to have a length of 1.45 m and a mass of 0.00338 kg. You are then asked to determine the following.
(a) amplitude
mm

(b) frequency
Hz

(c) wavelength
m

(d) wave speed
m/s

(e) direction the wave is traveling

• +x-direction
• -x-direction    

(f) tension in the rope
N

(g) average power transmitted by the wave
W

Answer 1

a) A = 2.35mm

b)w = 753 rad/sec

f = 753/2pi = 753/6.28 = 119.9 Hz

c)k=7 therefore

lamda = 6.28/7 =0.897 m

d)v = w/k = 753/7 = 107.57 m/sec

e)-ve x direction

f)T = v^2m/l = (107.57^2)*0.00338/1.45 = 26.97 N

g)Avg P = 0.5*m*v*w^2*A^2/l = 0.5*0.00338*107.57*753^2*(2.35*10^-3)^2/1.45 = 0.392 kg m^2/s^3

Answer 2

The wave function of a traveling wave on a thin rope is

y(x,t) = 2.35 mm cos[(7.00rad/m)x + (753 rad/s)t].

Comparin g it with standard equation

Y= Acos[kx+wt],we get

[a] Amplitude= 2.35mm

[b] frequency, f= w/2* pie= 753/2*3.14= 119.9 Hz

[c] wavelength, ?=2* pie/K= 2*3.14/7= 0.897 m

[d] wave speed, v=f* ?= 119.9*0.897= 107.6 m/s

[e] direction of wave travelling in ⁺x direction

[f] Tension

we have the expression, $v=\sqrt{\frac{T}{\mu }}$

?= m/l=0.00338 kg/1.45m = 0.00234 kg/m

T=v²*?=107.6²*0.00234= 27.05N

[g] Average power transmitted by the wave=(1/2)m*v*w²*A²

Pavg=0.5*0.00338* 107.6*753²*[2.35*10^-3]² = 0.64W