Instaneous Power in a standing Wave

In summary: Just use the following equation to find the average power over a complete cycle: \frac{P_a_v}{2} = \omega A_S_W TFMIn summary, the average power is zero.
  • #1
TFM
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0

Homework Statement



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

[tex] P(x,t) = =F\frac{\partial y(x,t)}{\partial x} \frac{\partial y(x,t)}{\partial t}[/tex]

where F is the tension.

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

Homework Equations



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

^ Power equation calculated from part 1

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
 
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  • #2
Does anyone have any idea what the best way is to approach this question?

Would be grearly appreciated,

TFM
 
  • #3
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
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
Given the instantaneous power, find the time average. What's the average value of [itex]\sin\omega t[/itex] or [itex]\cos\omega t[/itex] over a complete cycle?
 
  • #6
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
I don't quite understand your instantaneous power equation. Can you check for typos? If you can write it as [itex]P(x,t) = f(x)\cos\omega t[/itex], 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
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:

[tex] p_a_v_g = -F*(0) [/tex]
 
Last edited:
  • #9
TFM said:
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.
You cannot independently average different factors. You must average the entire function of t as a whole. You can show that [itex]\sin\omega t \cos\omega t[/itex] can be written as a simple sine function and thus its average must be zero.

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.
Your equation in post #1 as you wrote it here:
TFM said:
[tex] P(x,t) = -F sin(A_S_W(\omega t)kA_S_Wcos(kx))(A_S_Wsin(kx)\omega cos(\omega t)) [/tex]
doesn't quite make sense to me. (You've got some parentheses mixed up or something.)
 
  • #10
The equation from Mastering Physics:

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

TFM
 
  • #11
That's better.

Now you can write that equation as [itex]P(x,t) = f(x) \sin\omega t \cos\omega t[/itex]. Then show that the sine*cosine factor is equivalent to a simple sine function and thus averages to zero.
 
  • #12
Rearranging the equation, then:

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

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

TFM
 
  • #13
TFM said:
do I need to use doubke angle functions to show that the sine*cosine factor is equivalent to a simple sine function?
That's what I would do. :wink:
 
  • #14
Should I use:

[tex] sin2\theta = 2sin\theta cos\theta [/tex]

rearranging:

[tex] \frac{sin2\theta }{2} = sin\theta cos\theta [/tex]

Inserting into the equation, giving:

[tex] P(x,t) = \frac{sin2\theta }{2} F_x [/tex]

where [tex] F_x [/tex] is the other half of the equation

TFM
 
  • #15
I meant

[tex] P(x,t) = \frac{sin2\theta }{2} F_x [/tex]

TFM

(I tried editting, but it didn't change)
 
  • #16
Looks good, but keep it as [itex]\sin 2\omega t[/itex].
 
  • #17
[tex] P(x,t) = \frac{sin2\omega t}{2} F_x [/tex]

What would be the best thing to do now?

TFM
 
  • #18
Declare victory! Assuming that you accept that the average value of sin(t) = 0, you're done.
 

1. What is instantaneous power in a standing wave?

Instantaneous power in a standing wave is the power at a given point in time in a standing wave. It is the product of the instantaneous voltage and current at that point.

2. How is instantaneous power calculated in a standing wave?

Instantaneous power in a standing wave is calculated by multiplying the instantaneous voltage and current at a given point. This can be represented by the equation P(t) = V(t) x I(t).

3. How does instantaneous power vary in a standing wave?

Instantaneous power in a standing wave varies sinusoidally with time, as both the voltage and current vary sinusoidally. The maximum value of instantaneous power occurs when both voltage and current are at their maximum values, and the minimum value occurs when both are at their minimum values.

4. What is the relationship between instantaneous power and average power in a standing wave?

The average power in a standing wave is equal to the average of the instantaneous power over a period of time. This is because the positive and negative values of instantaneous power cancel out over a full cycle, resulting in an average value.

5. How can instantaneous power in a standing wave be controlled or manipulated?

Instantaneous power in a standing wave can be controlled and manipulated by adjusting the voltage or current at a given point. This can be achieved by changing the source voltage or by using devices such as capacitors and inductors to modify the current at specific points in the standing wave.

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