1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wave on a wire

  1. Mar 8, 2008 #1

    ~christina~

    User Avatar
    Gold Member

    [SOLVED] Wave on a wire

    1. The problem statement, all variables and given/known data
    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 [tex]y(x,t)= Asin( \omega t- kx) [/tex] 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

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

    3. The attempt at a solution

    a)
    [tex] f= 1/T= \omega/ 2 \pi [/tex]

    [tex]T= 0.002s[/tex]

    [tex] v= \omega/ k [/tex]

    [tex]2 \pi (500Hz)= \omega [/tex]

    [tex] \omega= 3151.59rad/s [/tex]

    [tex]k= \omega / v [/tex]

    [tex]3141.59rad/s / 196m/s= 16.02[/tex] => 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
     
  2. jcsd
  3. Mar 8, 2008 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    [tex] v= \sqrt{\frac{T}{\rho}} [/tex]

    For part c you would just differentiate wrt time.
     
  4. Mar 8, 2008 #3

    ~christina~

    User Avatar
    Gold Member

    Oh okay.
    Thanks alot Kurdt :smile:
     
  5. Mar 9, 2008 #4

    ~christina~

    User Avatar
    Gold Member

    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

    [tex] y(t)= A cos(\omega*t + \theta) [/tex]

    [tex]v(t)= y'(t)= -\omega A sin(\omega*t + \theta)[/tex]

    and

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

    But what would it be with 2 variables??

    would it be

    [tex] y(x/t)= (8x -8a)((Asin (omega*t- kx)) ? => (for here sinc they start with sin
    Thank you Kurdt
     
    Last edited: Mar 9, 2008
  6. Mar 9, 2008 #5

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  7. Mar 9, 2008 #6

    ~christina~

    User Avatar
    Gold Member

    alright. but does this apply always when they ask you these questions? do I just differentiate partially and then plug in?
     
  8. Mar 9, 2008 #7

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, that is the only way to find the transverse velocity and acceleration when you're given the wavefunction.
     
  9. Mar 9, 2008 #8

    ~christina~

    User Avatar
    Gold Member

    Thank you Kurdt :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Wave on a wire
Loading...