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

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In summary, the first graph shows the motion of a point at x=0 m, while the second graph shows the motion of the string at a time of 2 s. The amplitude of the oscillation is 1.0 m, the angular frequency is 2.90 rad/s, and the wavelength is 6.28 m. Questions regarding the vertical position of the string at x=0 m and t=0 s, the phase constant, the vertical displacement of the string at x=0 m and t=2 s, and the speed of the wave were discussed, with clarification on the use of symbols and equations provided.
  • #1
J6204
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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
 
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  • #2
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## ...
 
  • #3
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?
 
  • #4
yes, for example at ##t=2## and ##x=6## we have ##y=-1/2## ...
 
  • #5
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?
 
  • #6
you know ##\omega## and ##k## ...
 
  • #7
Ssnow said:
you know ##\omega## and ##k## ...
so then -1/2=1sin(6(2.90)-(6.28)(2)+φ), that seems to be correct
 
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  • #8
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!
 
  • #9
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
 

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

What is the meaning of "y(x,t)=A Sin(kx - wt + φ)"?

"y(x,t)=A Sin(kx - wt + φ)" is an equation that represents a sinusoidal wave, where A is the amplitude, k is the wave number, w is the angular frequency, x is the position, t is the time, and φ is the phase shift.

How do you interpret the graph of "y(x,t)=A Sin(kx - wt + φ)"?

The graph of "y(x,t)=A Sin(kx - wt + φ)" shows the displacement or height of the wave at a specific position x and time t. The amplitude A determines the maximum displacement, while the wave number k and angular frequency w affect the shape and period of the wave. The phase shift φ determines the horizontal shift of the wave.

What is the significance of the amplitude A in "y(x,t)=A Sin(kx - wt + φ)"?

The amplitude A represents the maximum displacement of the wave from its equilibrium position. It is a measure of the wave's energy and determines the intensity of the wave.

How do the wave number k and angular frequency w affect the graph of "y(x,t)=A Sin(kx - wt + φ)"?

The wave number k determines the number of complete cycles in a given distance, while the angular frequency w determines the number of complete cycles in a given time. These parameters affect the shape and period of the wave. A larger wave number or angular frequency results in a shorter wavelength and higher frequency wave.

What does the phase shift φ represent in "y(x,t)=A Sin(kx - wt + φ)"?

The phase shift φ represents the horizontal shift of the wave. It determines how much the wave is shifted to the left or right on the x-axis. A positive phase shift means the wave is shifted to the left, while a negative phase shift means it is shifted to the right.

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