How Do You Write the Function for a Traveling Wave on a String?

[Renaldo]

Homework Statement

A wave travels along a string in the positive x-direction at 27 m/s. The frequency of the wave is 46 Hz. At x = 0 and t = 0, the wave velocity is 2.6 m/s and the vertical displacement is y = 3 mm. Write the function y(x, t) for the wave. (Use the following as necessary: x and t. Assume x and y are in m, and that t is in s.)

Homework Equations

y(x,t) = Asin(κx - ωt + ∂)

ω = 2pif = 2pi(46) = 289.03

κ = ω/v = 289.03/27 = 10.70

The Attempt at a Solution

I determined that:

y(x,t) = Asin(10.70x - 289.03t + ∂)
y(0,0) = Asin(∂) = 0.003 m

Here is where I have trouble. I tried to take a partial derivative of the first function in order to solve for the velocity, since velocity is given. I've never encountered partial derivatives before, so I am not sure if what I did is correct. It doesn't give me the right answer.

y^t= -289.03Acos(10.70x - 289.03t + ∂)

If (x,t) = (0,0)
y^t = -289Acos(∂) = 2.6 m/s

I tried to combine this equation with

Asin(∂) = 0.003 m

to solve for ∂ and A, but I wasn't close to the final answer.

Is this a matter of taking the partial derivative wrong, or am I making some other mistake?

 

[Simon Bridge]

the wave velocity is 2.6 m/s
... the wave velocity is the speed of the wave in the x direction.

the partial derivative: ∂y(x,t)/∂t is the speed of point x in the y direction.

If the wave at t=0 has form $y(x,0)=f(x)$ and it is traveling in the +x direction with speed c,
then at time t it will have form: $y(x,t)=f(x-ct)$.

In your case, f(x) is the wave equation at t=0.

 

[Renaldo]

If the wave at t=0 has form $y(x,0)=f(x)$ and it is traveling in the +x direction with speed c, then at time t it will have form: $y(x,t)=f(x-ct)$.

In your case, f(x) is the wave equation at t=0.

Thanks for the response.

At t = 0, x = 0.

y(0,0) = Asin(∂)

According to the problem, at this point, y(0,0) = 0.003 m.

So:

Asin(∂) = 0.003 m.

I now have two variables, A and ∂. How do I solve for them?

 

[ehild]

y(x,t) = Asin(10.70x - 289.03t + ∂)
y(0,0) = Asin(∂) = 0.003 m

Here is where I have trouble. I tried to take a partial derivative of the first function in order to solve for the velocity, since velocity is given. I've never encountered partial derivatives before, so I am not sure if what I did is correct. It doesn't give me the right answer.

y^t= -289.03Acos(10.70x - 289.03t + ∂)

If (x,t) = (0,0)
y^t = -289Acos(∂) = 2.6 m/s

I tried to combine this equation with

Asin(∂) = 0.003 m

to solve for ∂ and A, but I wasn't close to the final answer.

Is this a matter of taking the partial derivative wrong, or am I making some other mistake?

You did it right. Show your further work.

ehild

 

[Simon Bridge]

Looking at it again and squinting this time:

At x = 0 and t = 0, the wave velocity is 2.6 m/s and the vertical displacement is y = 3 mm.

...this could be interpreted either way ... this wave velocity could be the rate of change of displacement y of whatever is at point x or it could be the speed of the whole wave-form (and if the latter, why mention the x position?) ... but it has already been said that:

A wave travels along a string in the positive x-direction at 27 m/s.

... OK, so c=27m/s and v_y(0,0) = 2.6m/s

So what was the final answer supposed to be?
What did you get?

 

[Renaldo]

Equation #1: -289Acos(∂) = 2.6 m/s
Equation #2: Asin(∂) = 0.003 m

Equation #1 (solved for A): A = 2.6/[-289.03cos(∂)]

Equation #2 (with substitution): (-2.6/289.03)tan(∂) = 0.003
(cont.) tan(∂) = -(0.003*289.03)/2.6
∂ = tan^-1[-(0.003*289.03)/2.6]
∂ = -0.32

Back to Equation #2: Asin(-0.32) = 0.003
A = -0.0095 (Amplitude is always positive, so can I just disregard the negative sign?)

The correct answer is: 0.0095 sin (10.70x - 289.03t + 2.82)

My answer as it is currently: 0.0095 sin (10.70x - 289.03t - 0.32)

My phase does not equal 2.82. Yet, using -0.32 as the phase resulted in the correct amplitude (assuming that it is alright to disregard the negative sign.)

 

[ehild]

Equation #1: -289Acos(∂) = 2.6 m/s
Equation #2: Asin(∂) = 0.003 m

Equation #1 (solved for A): A = 2.6/[-289.03cos(∂)]

Equation #2 (with substitution): (-2.6/289.03)tan(∂) = 0.003
(cont.) tan(∂) = -(0.003*289.03)/2.6
∂ = tan^-1[-(0.003*289.03)/2.6]
∂ = -0.32

Back to Equation #2: Asin(-0.32) = 0.003
A = -0.0095 (Amplitude is always positive, so can I just disregard the negative sign?)

The correct answer is: 0.0095 sin (10.70x - 289.03t + 2.82)

My answer as it is currently: 0.0095 sin (10.70x - 289.03t - 0.32)

My phase does not equal 2.82. Yet, using -0.32 as the phase resulted in the correct amplitude (assuming that it is alright to disregard the negative sign.)

You got A correctly. The easier way to calculate it is:

Acos(δ)=-2,6/289=-0.009
Asin(δ)=0.003

Square both equations and add them together: A²cos²δ+A²sin²δ=A²=90*10-6, A=0.0095.

From the first equations you see that sin(δ) >0 and cos(δ)<0. So δ is in which quadrant?

ehild

 

[Renaldo]

I see now. It is in the second quadrant. I need to add pi to -0.32 which results in 2.82.

Thanks! I think I can go to bed now...

 

[ehild]

Good night! :)

ehild