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

[J6204]

Homework Statement

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

 

[Ssnow]

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## ...

 

[J6204]

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?

 

[Ssnow]

yes, for example at ##t=2## and ##x=6## we have ##y=-1/2## ...

 

[J6204]

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?

 

[Ssnow]

you know ##\omega## and ##k## ...

 

[J6204]

you know ##\omega## and ##k## ...

so then -1/2=1sin(6(2.90)-(6.28)(2)+φ), that seems to be correct

 

[J6204]

you know ##\omega## and ##k## ...

I am not quite sure how to solve for φ in that equation, but thanks for your help anyways!

 

[Ssnow]

It is a goniometric equation ...
Ssnow

 

[J6204]

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

 

[J6204]

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

 

[J6204]

-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

 

[vela]

Are you sure about your value for the angular frequency? It looks closer to something like 2.1 rad/s to me.

 

[J6204]

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

 

[vela]

Oops, I calculated the period, not the frequency.

You just said y=-0.25 m. Did you mean y=-0.50 m?

 

[J6204]

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,

 

[vela]

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.

 

[J6204]

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?

 

[vela]

The way you tried was right. You're probably just plugging in the wrong numbers or messing up on the calculator.

 

[J6204]

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?

 

[vela]

It looks like you're multiplying ##\omega## by ##x##, which doesn't make sense. Also your value for ##k## is wrong.

 

[J6204]

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?

 

[vela]

Nope. A good place to start is to make sure you know what the symbols represent.

 

[J6204]

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

 

[J6204]

Nope. A good place to start is to make sure you know what the symbols represent.

Apparently 0.687 rad is incorrect also

 

[vela]

You can always check your answer by using the function you come up with to try reproduce the graphs you were given.

 

[J6204]

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