How is Wave Intensity Affected by Half Amplitude and Double Frequency?

In summary, the conversation discusses finding the intensity of a wave and the relationship between intensity, amplitude, and frequency. The questioner is unsure of the equation that relates intensity to amplitude and frequency and is asking for clarification. They also mention that the question is part of a proof and that they have seen the answer already. The conversation also addresses the importance of writing clearly and understanding context when discussing physics concepts.
  • #1
homeworkhelpls
41
1
TL;DR Summary: How do i find the intensity of this wave?

I know I is proportional to amplitude / frequency squared, but I don't know what equation this comes from. And I don't know how to answer this.
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  • #2
homeworkhelpls said:
I know I is proportional to amplitude / frequency squared
Let's start with the basics: do you know the difference between amplitude and frequency?
 
  • #3
Ibix said:
Let's start with the basics: do you know the difference between amplitude and frequency?
Yes, brother im an undergrad
 
  • #4
What kind of wave is this? And what is x on the vertical axis? Is it linear displacement of the particles in the wave?
 
  • #5
nasu said:
What kind of wave is this? And what is x on the vertical axis? Is it linear displacement of the particles in the wave?
doesnt matter, the question is used to ultimately prove that intensity of both waves is the same, its a proof question
 
  • #6
homeworkhelpls said:
Yes, brother im an undergrad
Ok. So is intensity proportional to the square of the amplitude or of the frequency, or what?
 
Last edited:
  • #8
@homeworkhelpls, you might be puzzled by these responses:
Ibix said:
Let's start with the basics: do you know the difference between amplitude and frequency?
Ibix said:
Ok. So is intensity proportional to the square of the amplitude or of the frequency, or what?
But I'd say the cause is some ambiguity in your post:
homeworkhelpls said:
I know I is proportional to amplitude / frequency squared,
I can read that as:
  • ##(\frac{amplitude}{frequency})^2##
  • either ##{amplitude}^2## or ##{frequency}^2##, but not sure which or whether it matters
  • both ##{amplitude}^2## and ##{frequency}^2##
My guess is that you meant the third, whereas @Ibix read it as the second.
 
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  • #9
haruspex said:
My guess is that you meant the third, whereas @Ibix read it as the second.
There's a point here about writing clearly, @homeworkhelpls . You saved yourself two characters by writing / writing of "and". You'd probably have had your answer this morning if you'd spent the two extra characters.
 
  • #10
homeworkhelpls said:
doesnt matter, the question is used to ultimately prove that intensity of both waves is the same, its a proof question
It does matter to know what quantities are represented on the axes. You may know it but out of context is not obvious.
 
  • #11
nasu said:
It does matter to know what quantities are represented on the axes. You may know it but out of context is not obvious.
No i don't know it, you're not supposed to know it bro i saw the answer
 
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  • #12
haruspex said:
@homeworkhelpls, you might be puzzled by these responses:But I'd say the cause is some ambiguity in your post:

I can read that as:
  • ##(\frac{amplitude}{frequency})^2##
  • either ##{amplitude}^2## or ##{frequency}^2##, but not sure which or whether it matters
  • both ##{amplitude}^2## and ##{frequency}^2##
My guess is that you meant the third, whereas @Ibix read it as the second.
Damn guys my bad lol, no all three guesses were wrong i meant the physics law of intensity where, I is directally proportional to amplitude squared
 
  • #13
kuruman said:
Maybe you will find this link useful.
lol, what if i have no heat source
 
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  • #14
homeworkhelpls said:
Damn guys my bad lol, no all three guesses were wrong i meant the physics law of intensity where, I is directally proportional to amplitude squared
See if this helps https://physics.info/intensity/
 
  • #15
homeworkhelpls said:
I know I is proportional to amplitude / frequency squared, but I don't know what equation this comes from. And I don't know how to answer this.
It is not clear what your difficulty is. You don’t need the equation to apply proportionality. For example: the area of a circle is 100cm²; the radius is halved; what is the new area? (No equations needed.)

From the graphs:

1: what is the value of ##\frac {A_P}{A_Q}## (where ##A## is amplitude)?

2: what is the value of ##\frac {f_P}{f_Q}## (where ##f## is frequency)?

In terms of proportionality:

3: how is intensity related to amplitude if all other parameters are kept constant?

4: how is intensity related to frequency, if all other parameters are kept constant?

If you can answer all 4 questions, you should be able to solve the problem.
 
  • #16
homeworkhelpls said:
No i don't know it, you're not supposed to know it bro i saw the answer
You know that this is related to some chapter in a book where a specific type of wave is described. Possibly just a plane wave in 1 dimension. But this is not the only wave possible and a linear dispalcement (position of particle) is not the only parameter used to describe a wave. What you know in the context of the specific book is not obvious for the people outside that context. A sound wave is most commonly described by the acoustic pressure and not particle displacement, for example.
 
  • #17
Bruh relax I’ll just send answer
nasu said:
You know that this is related to some chapter in a book where a specific type of wave is described. Possibly just a plane wave in 1 dimension. But this is not the only wave possible and a linear dispalcement (position of particle) is not the only parameter used to describe a wave. What you know in the context of the specific book is not obvious for the people outside that context. A sound wave is most commonly described by the acoustic pressure and not particle displacement, for example
 
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FAQ: How is Wave Intensity Affected by Half Amplitude and Double Frequency?

What is the intensity of a wave at half amplitude compared to the reference wave?

The intensity of a wave is proportional to the square of its amplitude. Therefore, if the amplitude is halved, the intensity is reduced to one-quarter (1/4) of the original intensity of the reference wave.

How does doubling the frequency of a wave affect its intensity?

Doubling the frequency of a wave does not directly affect its intensity, as intensity is primarily dependent on amplitude. However, in some contexts, higher frequency waves may interact differently with the medium, but this is not a direct consequence of the frequency change alone.

What is the combined effect on intensity when the amplitude is halved and the frequency is doubled?

When the amplitude is halved, the intensity becomes one-quarter (1/4) of the original intensity. Doubling the frequency does not change this result. Therefore, the combined effect results in the intensity being one-quarter (1/4) of the reference wave's intensity.

Why is intensity proportional to the square of the amplitude?

Intensity is proportional to the square of the amplitude because it represents the power per unit area carried by the wave. Since power is related to the energy transported by the wave, and energy is proportional to the square of the amplitude, intensity follows this relationship.

Does changing the frequency of a wave affect its speed?

The speed of a wave in a given medium is generally determined by the properties of the medium and is independent of the wave's frequency and amplitude. However, in dispersive media, different frequencies may travel at different speeds.

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