# Homework Help: Instaneous Power in a standing Wave

1. Feb 28, 2008

### TFM

1. The problem statement, all variables and given/known data

The instantaneous rate at which a wave transmits energy along a string (instantaneous power) is

$$P(x,t) = =F\frac{\partial y(x,t)}{\partial x} \frac{\partial y(x,t)}{\partial t}$$

where F is the tension.

Show that for all values of x, the average power $$P_a_v$$ carried by the standing wave is zero. (Equation $$P_a_v = \frac{1}{2}\sqrt{\mu F} \omega^2 A^2$$ does not apply here. Can you see why?)

2. Relevant equations

$$P(x,t) = -F sin(A_S_W(\omega t)kA_S_Wcos(kx))(A_S_Wsin(kx)\omega cos(\omega t))$$

^ Power equation calculated from part 1

3. The attempt at a solution

The answer section is in the form of a writing box (I am using Mastering Physics), Not the normal maths box.

What is the best way to show for all values of x the power is zero?

TFM

2. Feb 28, 2008

### TFM

Does anyone have any idea what the best way is to approach this question?

Would be grearly appreciated,

TFM

3. Feb 29, 2008

### TFM

This is the annoying part of Mastering Physics - I can't get onto the next question until I finish this one. What I would normally do for a "Show that for all values of x..." question, I would just stick in thousands of values for x.

As you can see, any ideas would be very greatly appreciated.

TFM

4. Feb 29, 2008

### TFM

What is the best way to go about this question, should I just use a lot o different values, or is there a way to change the equation into an average one?

ANy help would be greatly appreciated,

TFM

5. Feb 29, 2008

### Staff: Mentor

Given the instantaneous power, find the time average. What's the average value of $\sin\omega t$ or $\cos\omega t$ over a complete cycle?

6. Feb 29, 2008

### TFM

Since the maximam value is 1, the minimum is -1, the average will be 0.

Would you insert this intop the orginal power equation?

TFM

7. Feb 29, 2008

### Staff: Mentor

I don't quite understand your instantaneous power equation. Can you check for typos? If you can write it as $P(x,t) = f(x)\cos\omega t$, then you can immediately see that the average is zero since the average of sin or cos is zero (they are negative as much as positive).

8. Feb 29, 2008

### TFM

If cos(average) = 0, then it will make the average power 0 becasue all the terms are multiplied together, so the 0 from cos(average) and sin(average) will make the whole equation = 0.

The equation is right, because I calculated it for the first part oif the question, on Mastering Physics, and it came up as it correct.

TFM

edit:

It will make the formula:

$$p_a_v_g = -F*(0)$$

Last edited: Feb 29, 2008
9. Feb 29, 2008

### Staff: Mentor

You cannot independently average different factors. You must average the entire function of t as a whole. You can show that $\sin\omega t \cos\omega t$ can be written as a simple sine function and thus its average must be zero.

Your equation in post #1 as you wrote it here:
doesn't quite make sense to me. (You've got some parentheses mixed up or something.)

10. Feb 29, 2008

### TFM

The equation from Mastering Physics:

$$P(x,t) = -F (sin(\omega t)kA_S_Wcos(kx))(A_S_Wsin(kx) \omega cos(\omega t))$$

TFM

11. Feb 29, 2008

### Staff: Mentor

That's better.

Now you can write that equation as $P(x,t) = f(x) \sin\omega t \cos\omega t$. Then show that the sine*cosine factor is equivalent to a simple sine function and thus averages to zero.

12. Feb 29, 2008

### TFM

Rearranging the equation, then:

$$P(x,t) = sin(\omega t)cos(\omega t)[-F(kA_S_Wcos(kx))(\omega A_S_Wsin (kx))]$$

do I need to use doubke angle functions to show that the sine*cosine factor is equivalent to a simple sine function?

TFM

13. Feb 29, 2008

### Staff: Mentor

That's what I would do.

14. Feb 29, 2008

### TFM

Should I use:

$$sin2\theta = 2sin\theta cos\theta$$

rearranging:

$$\frac{sin2\theta }{2} = sin\theta cos\theta$$

Inserting into the equation, giving:

$$P(x,t) = \frac{sin2\theta }{2} F_x$$

where $$F_x$$ is the other half of the equation

TFM

15. Feb 29, 2008

### TFM

I meant

$$P(x,t) = \frac{sin2\theta }{2} F_x$$

TFM

(I tried editting, but it didn't change)

16. Feb 29, 2008

### Staff: Mentor

Looks good, but keep it as $\sin 2\omega t$.

17. Feb 29, 2008

### TFM

$$P(x,t) = \frac{sin2\omega t}{2} F_x$$

What would be the best thing to do now?

TFM

18. Feb 29, 2008

### Staff: Mentor

Declare victory! Assuming that you accept that the average value of sin(t) = 0, you're done.