Analyzing a Wave: Homework Solutions

[aliaze1]

Homework Statement

http://photo.ringo.com/230/230997145O975467146.jpg

http://photo.ringo.com/230/230997145O975467146.jpg

Homework Equations

λ=v/f
k=2π/λ
ω=2πf
T=1/f
v=ω/k
V(string) = √(Tension of string/μ), where μ = denisty

D(x,t) = A sin (kx - ωt + Φ)

The Attempt at a Solution

I found the maximum displacement as 2, found from the given equation

The third part seemed to be the next simplest, so using v=ω/k, I calculated 638/12.57 as the speed, which was incorrect

To my knowledge, this speed is needed to calculate the tension of part 1

 

[Astronuc]

By any chance, is the third question asking for the lateral velocity (dy/dt) of a point on a string rather than the wave velocity?


See the relationship of wave velocity and tension here
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html

 

[aliaze1]

i guess it is possible, but that is the exact question copied word-for-word

 

[rootX]

what does d(D(x,y))/dt means?

 

[aliaze1]

I just realized that the third part is not required to complete the first part,

using v=ω/k, and plugging this v into

V(string) = √(Tension of string/μ), where μ = denisty;

I get an answer of 12880.7, which is essentially 12.9 *10³...the answer however is simply 12.9...

where am i going wrong?

thanks

 

[aliaze1]

I thought it had to do with the amplitude being in centimeters, so I divided by 100, but that still is 129 not 12.9...

 

[Astronuc]

Make sure all units are consistent. The linear mass density is 5 g/ m as opposed to 0.05 g /cm or 0.005 kg/m. Perhaps that is where one is off by 3 or 1 order of magnitude depending on the values one uses. Tension should be in Newtons (for SI/mks).

 

[aliaze1]

Make sure all units are consistent. The linear mass density is 5 g/ m as opposed to 0.05 g /cm or 0.005 kg/m. Perhaps that is where one is off by 3 or 1 order of magnitude depending on the values one uses. Tension should be in Newtons (for SI/mks).

thanks!, makes sense