A traveling wave on an infinite string, find the intensity

Click For Summary
SUMMARY

The discussion centers on calculating the intensity of a traveling wave on an infinite string, with the correct power output determined to be approximately 40 Watts. The user initially confused units of frequency (Hz) with power (Watts) and referenced the formula = 0.5((pλ)(ωA)²)v, where p is linear density, λ is wavelength, ω is angular frequency, A is amplitude, and v is phase speed. The conversation highlights the importance of correctly applying formulas and units in wave mechanics.

PREREQUISITES
  • Understanding of wave mechanics and intensity calculations
  • Familiarity with the concepts of linear density and impedance
  • Knowledge of angular frequency and amplitude in wave equations
  • Ability to manipulate and apply relevant equations in physics
NEXT STEPS
  • Study the derivation and application of the wave intensity formula = 0.5((pλ)(ωA)²)v
  • Learn about the relationship between tension, density, and wave speed using v = √(T/ρ)
  • Explore the concept of impedance in wave mechanics and its calculation
  • Investigate common mistakes in unit conversions and their impact on physical calculations
USEFUL FOR

Students and professionals in physics, particularly those studying wave mechanics, as well as educators looking to clarify concepts related to wave intensity and unit conversions.

LCSphysicist
Messages
644
Reaction score
162
Homework Statement
I will post below.
Relevant Equations
.
1598913360062.png

The answer is about 40 Watts, but i am really not getting it:
<Pot> = 0.5((pλ)(ωA)²)v

p density linar
λ wavelength
ω angular frequency
A amplitude
v phase speed

Why is this wrong?

(I already tried too by <Pot> = Z*(ωA)²/2)

Z is impedance
 

Attachments

  • 1598904170440.png
    1598904170440.png
    11.7 KB · Views: 143
Last edited:
Physics news on Phys.org
LCSphysicist said:
The answer is about 40 Hz
Hz is not a unit of power. Did you mean Watts?
 
  • Like
Likes LCSphysicist
berkeman said:
Hz is not a unit of power. Did you mean Watts?
Opsss, yes >.< i wrote wrong
 
  • Like
Likes berkeman
The Relevant Equations: has an error.
It shows '1 lb = 3[/color][/color]54 gm-wt', it should be 4[/color]54 gm-wt.[/color][/color][/color][/color]
 
  • Informative
Likes LCSphysicist
LCSphysicist said:
i am really not getting it:
Where is your working? How did you find the wavelength?
We cannot say what is wrong without seeing what you did.
 
  • Like
Likes LCSphysicist
haruspex said:
Where is your working? How did you find the wavelength?
We cannot say what is wrong without seeing what you did.
v = √T/ρ = λ*f is what i am using to find the speed
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
20
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
Replies
5
Views
4K
Replies
9
Views
2K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K