Graphs of y(x,t)=A Sin(kx - wt + φ) are shown below

  • Thread starter Thread starter J6204
  • Start date Start date
  • Tags Tags
    Graphs
AI Thread Summary
The discussion revolves around solving equations related to wave motion represented by the function y(x,t)=A Sin(kx - wt + φ). Key calculations include determining the amplitude, angular frequency, and wavelength, with values found to be 1.0m, 2.90 rad/s, and 6.28m respectively. Participants express confusion about calculating the vertical position of the string at specific points and the phase constant, φ. There is a focus on using given graphs to deduce values, with attempts to solve for φ leading to various incorrect results. The conversation emphasizes the importance of correctly interpreting wave parameters and suggests verifying answers by reproducing the original graphs.
J6204
Messages
56
Reaction score
2

Homework Statement


graphs.png
The first graph shows y vs t for a point at x=0 m. The second shows y vs x for the string at a time of 2 s.

I have calculated the amplitude of the oscillation to be 1.0m

I have calculated the angular frequency of the oscillation to be 2.90 rad/s

I have calculated the wavelength of the oscillation to be 6.28m

However the following questions I am not quite sure how to do based on the graphs I have linked in the question

1) Calculate the vertical position of the string at x=0m, t=0s.

2) Calculate the phase constant, φ, of the motion.

3) Determine the vertical displacement of the string element at x=0m at time t=2s.

4) What is the speed of the wave?

Homework Equations

The Attempt at a Solution


1) if we let x=0m and t=0m then y=1m sin φ but how do I solve for the vertical displacement if there are two unknowns?

3) similar to number 1, again not quite sure how to solve for the vertical displacement

2), 4) not quite sure how to do these either
 
Physics news on Phys.org
For the ##1)## you must have solved ##2)##. But you know ##\omega## and ##k## and selecting a precise point ##(x_{0},t_{0})## on the graphs you can deduce ##\varphi## ...
 
Ssnow said:
For the ##1)## you must have solved ##2)##. But you know ##\omega## and ##k## and selecting a precise point ##(x_{0},t_{0})## on the graphs you can deduce ##\varphi## ...
So for number 2 I can pick any point?
 
yes, for example at ##t=2## and ##x=6## we have ##y=-1/2## ...
 
Ssnow said:
yes, for example at ##t=2## and ##x=6## we have ##y=-1/2## ...
okay so then -1/2=1m sin Sin(6k - 2w + φ), so then how do I solve for φ?
 
you know ##\omega## and ##k## ...
 
Ssnow said:
you know ##\omega## and ##k## ...
so then -1/2=1sin(6(2.90)-(6.28)(2)+φ), that seems to be correct
 
  • Like
Likes Ssnow
Ssnow said:
you know ##\omega## and ##k## ...
I am not quite sure how to solve for φ in that equation, but thanks for your help anyways!
 
It is a goniometric equation ...
Ssnow
 
  • #10
Ssnow said:
It is a goniometric equation ...
Ssnow
-1/2=sin(4.84+φ) so then what is φ? I have never solved a equation like this before
 
  • #11
Ssnow said:
It is a goniometric equation ...
Ssnow
I tried to solve 6*2.90-6.28*2+a=-pi/6, which gave a=-5.36rad but this was incorrect
 
  • #12
J6204 said:
-1/2=sin(4.84+φ) so then what is φ? I have never solved a equation like this before
I also tried 2.53rad but this was incorrect
 
  • #13
Are you sure about your value for the angular frequency? It looks closer to something like 2.1 rad/s to me.
 
  • #14
vela said:
Are you sure about your value for the angular frequency? It looks closer to something like 2.1 rad/s to me.
2.90rad/s was correct for the angular fequency, y=-1/2 when x=6 but when t=2 it looks like it is y=-.250, as was stated above to use. I have been stuck on this question for quite some time and am trying to finish it off
 
  • #15
Oops, I calculated the period, not the frequency.

You just said y=-0.25 m. Did you mean y=-0.50 m?
 
  • #16
vela said:
Oops, I calculated the period, not the frequency.

You just said y=-0.25 m. Did you mean y=-0.50 m?
Well on the first graph y vs t when t is 2 it doesn't quite look like y=-0.5, however on the second graph y vs x when x is 6 y=-0.5,
 
  • #17
The first graph is of the motion of the point x=0 on the wave. If you use y=-0.25 m for t=2 s, you need to set x=0 m.
 
  • #18
vela said:
The first graph is of the motion of the point x=0 on the wave. If you use y=-0.25 m for t=2 s, you need to set x=0 m.
Oh okay. So would you be able to help me solve the phase constant then?
 
  • #19
The way you tried was right. You're probably just plugging in the wrong numbers or messing up on the calculator.
 
  • #20
vela said:
The way you tried was right. You're probably just plugging in the wrong numbers or messing up on the calculator.
I did 7pi/6= (6×2.90−6.28×2+a) solved for a and got 2 numbers both in degrees, 145, -35 I changed 145 degrees to radians and got 2.53. What did I do wrong?
 
  • #21
It looks like you're multiplying ##\omega## by ##x##, which doesn't make sense. Also your value for ##k## is wrong.
 
  • #22
vela said:
It looks like you're multiplying ##\omega## by ##x##, which doesn't make sense. Also your value for ##k## is wrong.
Isn't w=6.28 and k=2.90?
 
  • #23
Nope. A good place to start is to make sure you know what the symbols represent.
 
  • #24
vela said:
Nope. A good place to start is to make sure you know what the symbols represent.
Yes you are right, so k=2π/λ, then k= 2π/6.28, k = 1. Therefore the equation I should be solving is 7pi/6=1*2-6*6.28+a which gives a=39.35 degrees which is 0.687 rad for the phase constant? Please let that be right lol
 
  • #25
vela said:
Nope. A good place to start is to make sure you know what the symbols represent.
Apparently 0.687 rad is incorrect also
 
  • #26
You can always check your answer by using the function you come up with to try reproduce the graphs you were given.
 
  • #27
vela said:
You can always check your answer by using the function you come up with to try reproduce the graphs you were given.
Do you know what the phase fondant works out to be? Once I can do that I can do the other 2 questions
 
Back
Top