Wave on a Wire: Finding Equations and Tension

[~christina~]

[SOLVED] Wave on a wire

Homework Statement

A transverse wave on a taut wire has amplitude of 0.200mm and a frequency of 500Hz. It travels with a speed of 196m/s.

a) write an equation in SI units of the form y(x,t)= Asin( \omega t- kx) for this wave

b)The linear mass density of this wire is 4.10g/m Find the Tension in the wire

c) what are the transverse velocity and the acceleration of the wave when x= 19.7m and t= 0.101s

Homework Equations

F= -kx
v= \omega/ k ==> teacher gave me this equation but I can't find it in the book...is it valid?
f= 1/T= \omega/ 2 \pi

The Attempt at a Solution

a)
f= 1/T= \omega/ 2 \pi

T= 0.002s

v= \omega/ k

2 \pi (500Hz)= \omega

\omega= 3151.59rad/s

k= \omega / v

3141.59rad/s / 196m/s= 16.02 => I'm not sure it's suppsosed to be that large

I guess I'd just plug in the numbers but I'm not sure if the way I got the numbers is correct.

b) I don't know how to find this

c) I think I would just differentiate the original equaiton with the numbers included and then just plug in the values given and find the numbers.

I have a question though.

Is the transverse velocity always found through the the differential equation?


Thank you very much

 

[Kurdt]

Part a looks ok. For part b, the speed of a wave on a string with tension T and mass per unit length \rho is:

v= \sqrt{\frac{T}{\rho}}

For part c you would just differentiate wrt time.

 

[~christina~]

Part a looks ok. For part b, the speed of a wave on a string with tension T and mass per unit length \rho is:

v= \sqrt{\frac{T}{\rho}}

For part c you would just differentiate wrt time.

Oh okay.
Thanks a lot Kurdt

 

[~christina~]

For part c you would just differentiate wrt time.

I have a Question about this don't I have to differentiate with respect to BOTH time and displacement?.. the problem said that I had to find the transverse veloicty and acceleration of the wave when x= 19.7m and t= 0.101s so how would I do this?

I know that I differentiate but I'm not sure how it would look if I differentiate with both x and t together...

I do know that if it's just t and theta then it would be

y(t)= A cos(\omega*t + \theta)

v(t)= y'(t)= -\omega A sin(\omega*t + \theta)

and

a(t)= v'(t)= -\omega^2 A cos(\omega t + \theta)

But what would it be with 2 variables??

would it be

y(x/t)= (8x -8a)((Asin (omega*t- kx)) ? =&gt; (for here sinc they start with sin<br /> Thank you Kurdt

 

[Kurdt]

You just differentiate wrt time. You treat the x variable as a constant when you do this. Then once you've differentiated you plug in the numbers.

 

[~christina~]

You just differentiate wrt time. You treat the x variable as a constant when you do this. Then once you've differentiated you plug in the numbers.

alright. but does this apply always when they ask you these questions? do I just differentiate partially and then plug in?

 

[Kurdt]

alright. but does this apply always when they ask you these questions? do I just differentiate partially and then plug in?

Yes, that is the only way to find the transverse velocity and acceleration when you're given the wavefunction.

 

[~christina~]

Yes, that is the only way to find the transverse velocity and acceleration when you're given the wavefunction.

Thank you Kurdt