Determine the amplitude of a mechanical wave given only it's velocity and period

[staticd]

Homework Statement

A transverse sinusiodal wave on a string has a period T= 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t=0, an element of the string at x=0 has a transverse position of 2.00 cm and is traveling downward with a speed of 2.00 m/s.

Homework Equations

I know that the general equation for a waveform is:

y(x,t)= Asin(kx-(omega)t+(phi))

{omega is the angular frequency and phi is the phase constant}

I also know that at t=0 and x=0

y(x,t)=.02 m --> .02=Asin(phi) --> A=.02/sin(phi)

I can also determine omega given the period:

omega=(2*pi)/.025=251.3274 rad/s

The Attempt at a Solution

Even with my relevant equations, I cannot figure out what I am supposed to do with the extra known factor, velocity. I have scrubbed the current chapter and can't find where I can use the transverse position, angular frequency and velocity in the y direction to determine a waves amplitude. What am I missing? Please help! I have so many problems to do but I just can't do another one until I figure this out! Thanks so much.

 

[Zaphys]

There`s certainly a way of getting A with that data. I´ll be glad to explain it to you but it's a little heavy, I mean all the calculations.

Let's start:

we are given the constants of the wave T which, as you said, give us w (omega) and v which give us k (by the relationship v=(lamda)/T and from that k=2·(pi)/(lamda) ). We also have the elongation y(0,0)=Yo (i know the data is numerical but in order to have it more simple we´ll work with letters) and the velociti of that point at that time Vy(0,0)=Vo. (I think this is the fact that you misregarded in your first post)

Ok then we can write the eq. like this (as you did):

y(x,t)=A·sin(kx-wt+&) [& is (phi)]

the first condition we know is y(0,0)=Yo , let's write it:

y(0,0)=A·sin(k·0-w·0+&)=A·sin&=Yo -> A·sin&=Yo

first relationship between unknown & and A.

Now we use the fact that velocity of the point x=0 at t=0 is 2m/s downwards which makes:

Vo=-2 m/s so we´ll work with Vo.

You should remember or know that the velocity of a point at a moment is given by the partial derivative with respecto to time of the wave function y(x,t)=... So let's differenciate:

[tex]\partial[/tex]y/[tex]\partial[/tex]t=-Aw·cos(kx-wt+&)

so,

Vy(0,0)=-Aw·cos(K·0-w·0+&)=-Aw·cos&=Vo -> -Aw·cos&=Vo

Second relationship between unkanown A and &.

Now you just have to solve the eqs. system:

A·sin&=Yo

A·w·cos&=Vo

 

[Zaphys]

My personal advice with the system is to divide first eq. by the other and write the new eq. (without A) like this:

sin&=(w·Yo/Vo)·cos&

And the use the trigonometric relationship sin²&+cos²&=1 in order to get the expression of sin& (as we want A, not &) then once you've done all the calculations just put the expression you get for sin& in the one of the first eqs. which had sin& on it:

A·sin&=Yo -> A=Yo/sin&

And there you have your solution.

Hope it is worth and hope I made myself clear. If you have any other trouble just tell me.

Salutations, zaphys. :)

 

[rl.bhat]

When a transverse sinusoidal wave travels on a string, every element of the string executes simple harmonic motion. Its equation is given by
y = A*sin(ωt), where y is the transverse displacement, A is the amplitude and ω is the angular velocity = 2π/T.
The transverse velocity is given by
dy/dt = A*ω*cos(ωt) = ω*sqrt[ A^2 - A^2*sin^2(ωt)] = ω*sqrt[ A^2 - y^2]
Angular velocity, transverse velocity and transverse displacement is given. Find the wave amplitude.

 

[staticd]

Why didn't I see that? Thank you so much...(solving for A reveals the amplitude to be 2.15 cm).

I think I just hit the wall in my head where I am not thinking holistically about the problem.

Thanks again.

 

[staticd]

Wait a second, rl.

How did you come up with that equality for the derivative of the waveform?

I think that was what I was actually looking for!

 

[rl.bhat]

Wait a second, rl.

How did you come up with that equality for the derivative of the waveform?

I think that was what I was actually looking for!

If y is the displacement, then dy/dt will be the velocity. Using simple trigonometry I got the equality.

 

[staticd]

If y is the displacement, then dy/dt will be the velocity. Using simple trigonometry I got the equality.

Thank you so much for your help with this. While I did find the solution, I am having trouble visualizing what this means. Do you know, or can you draw, the graphical representation of how to solve for A? Many thanks.

 

[staticd]

guess i get no love with this request---

 

[rl.bhat]

guess i get no love with this request---

Sorry. I don't how to do that.

 

[staticd]

No worries! In my mind, I picture a circle that has a radius equal to A. Where the angular frequency is the rate at which A fluctuates from +A to -A in radians per second. I am just not sure what that really looks like or if I am way of base. Your equation says that if I know the transverse position and the angular frequency then I can determine the amplitude. I am just trying to figure out how this all ties together because I think it's an important fundamental concept. Thanks for the try.