Question

A transverse harmonic wave travels on a rope according to the following expression:
y(x,t) = 0.15sin(2.1x + 17.3t)

The mass density of the rope is μ = 0.125 kg/m. x and y are measured in meters and t in seconds.

1. What is the wavelength of the wave?
2. What is the speed of the wave?
3. What is the tension in the rope?
4. At x = 3.2 m and t = 0.4 s, what is the acceleration of the rope?
5. What is the average speed of the rope during one complete oscillation of the rope?

Answer 1

I have fount that the amplitude is:
\(A=0.15~\color{grey}{\rm m}\)

and the frequency of oscillation of the wave is:
\(f=2.75338~\color{grey}{\rm Hz}\)

Answer 2

I am stuck from there:( Anyone, help?

Answer 3

Oh, found the speed of the wave:
17.3 / 2.1 = 8.2381 m/s

Answer 4

Then, wavelength is:
f × λ = v
w/(2π) × λ = v
17.3/(2π) × λ = 8.2381 m/s
λ = 2.99199 m

Answer 5

How do I find the acceleration? Normally I would differentiate with respect to t, but here I have two variables.

Answer 6

Just about the same.... I just have a few more digits

Answer 7

\(y(x,t) = 0.15\sin(2.1x + 17.3t)\)

\(a(x,t)=\partial^2y/\partial t^2= -44.8935 \sin[17.3 t + 2.1 x]\)
\(a(3.2,0.4)=-39.4586\)

Answer 8

But I keep thinking about Tension and I seem to be unable to figure it out. ('m so bad at tensions)

Answer 9

Is there such a thing of conservation of Tension? (xD)
I mean that the Tension in the rope is constant at all time...

Answer 10

Oh, there found the equation
\(v = \sqrt{T/μ}\)
\(v^2 = T/μ\)
\( T=0.125\cdot 8.2381^2=8.48333~N\)

Answer 11

One more left for this set:
The average velocity. (For some reason when I type everything in here at least a part of my intellect comes back to me)

Answer 12

8.48333

Answer 13

:)

Answer 14

Looks good