Question about finding the amplitude of this circular wavefront

[randyrandy]

Homework Statement

A transverse circular wave propagates on a plane surface without energy dissipation. If the wave amplitude is 2 cm at a distance of 10 cm from its centre, what is the wave amplitude at a distance of 40 cm from its centre?

answer is 1cm.

Relevant Equations

speed = frequency*wavelength

I'm stumped. How do you get this without knowing the wavelength? Can someone explain how to get the answer?

 

[haruspex]

Homework Statement:: A transverse circular wave propagates on a plane surface without energy dissipation. If the wave amplitude is 2 cm at a distance of 10 cm from its centre, what is the wave amplitude at a distance of 40 cm from its centre?

answer is 1cm.
Relevant Equations:: speed = frequency*wavelength

I'm stumped. How do you get this without knowing the wavelength? Can someone explain how to get the answer?

What expressions do you have for the relationship between amplitude and energy or power?

 

[randyrandy]

Part of what threw me off was I was uncertain whether the amplitude of 2cm is the peak amplitude for the wave or some height between the maximum amplitude.

As to the power equation, I'm not sure. The problem didn't mention anything about power. Oh, wait, energy is conserved so this is really about energy distributed over a greater area which translates to amplitude? so at radius of 10 it's 2cm. so at radius of 40cm, there must be a ratio? something to do with pi*r^2? something drops off exponentially as you go further out from center. (sort of like sound or light that dims the further you are from the source like stars?) 100/1600 = x/2cm. but then the answer is 2/16 or 1/8th. not sure...

 

[haruspex]

about energy distributed over a greater area which translates to amplitude?

Think about a single circular wavefront. As it travels out it is distributed over a greater...?
The way amplitude relates to energy is crucial. What do you know about waves?

something drops off exponentially as you go further out

Not exponentially.

 

[randyrandy]

maybe circumference? rather than area, just the perimeter of the wave front? so it's pi*2*r? so linear? just intuitively, i know that a ripple will die down as the wave moves from the center...so the amplitude will be smaller...so it'll be inverse relationship to distance.

2*10cm = 1/2cm,
2*40cm = 1/xcm

10cm/40cm = x/2

x = .5cm

but that's not the answer...

 

[HallsofIvy]

Yes, the problem said this is a circular wave front in a plane surface. So the "wave front" is a circle and it length is proportional to the radius of the circle. 40 cm is 4 times as much as 10 cm so the wave front is 4 as long. Since the wave height is 2 cm at 10 cm radius, the energy will be spread over 4 times the length. The wave height at 40 cm will be 2/4= 1/2 cm.

 

[randyrandy]

But the answer says it's 1cm not 0.5cm...

 

[HallsofIvy]

But the answer says it's 1cm not 0.5cm...

Then that answer is wrong!

 

[PeroK]

But the answer says it's 1cm not 0.5cm...

How is energy related to amplitude?

 

[haruspex]

But the answer says it's 1cm not 0.5cm...

You overlooked this part of my post:

The way amplitude relates to energy is crucial. What do you know about waves?

Or think about SHM, which is essentially wave motion. How does energy in a spring relate to displacement?

 

[Cutter Ketch]

Then that answer is wrong!

No, I’m afraid it is right.

 

[randyrandy]

I'm confused more than ever...

 

[haruspex]

I'm confused more than ever...

Then I will ask you again

How does energy in a spring relate to displacement?

 

[randyrandy]

Then I will ask you again

1/2kx^2? = energy. but that formula doesn't have radius as a variable.

 

[randyrandy]

No, I’m afraid it is right.

Could you explain how to solve this problem?

 

[haruspex]

1/2kx^2? = energy. but that formula doesn't have radius as a variable.

If you recall, I was trying to get you to think about the relationship between energy and amplitude in waves. In waves and SHM, amplitude is max displacement from equilibrium.

 

[randyrandy]

If you recall, I was trying to get you to think about the relationship between energy and amplitude in waves. In waves and SHM, amplitude is max displacement from equilibrium.

I found this formula for a spring F=kx. Would that be applicable here?

 

[haruspex]

I found this formula for a spring F=kx. Would that be applicable here?

No, you already have the relevant spring formula, E=½kx². You just need to see what the max displacement of an oscillating spring corresponds to in wave motion.

 

[randyrandy]

No, you already have the relevant spring formula, E=½kx². You just need to see what the max displacement of an oscillating spring corresponds to in wave motion.

Would the problem assume that 2cm is the max displacement at 10cm from center? This seemed ambiguous to me.

therefore... 1/2k4 = energy (energy would be proportional to circumference pi*2*10cm) at 10cm. at 40cm... 1/2kx^2. what would the energy side of the equation look like at 40cm? pi*2*40cm?4/x^2 = 40/10 , x = 1cm.

holy crap.

wow! that was totally not intuitive. what am i missing in terms of connecting that pivotal spring formula to wave propagation?

formally speaking, what energy equation would be used on the other side of 1/2kx^2 = ? that uses radius in it's equation?

 

[haruspex]

Would the problem assume that 2cm is the max displacement at 10cm from center?

Yes, that's what wave amplitude means. It is the max displacement from the mean position.
For a surface wave on water, the mean position is where the surface would be if not for the wave, so the amplitude is the height of a wave crest above that.
For a longitudinal wave, such as an air column in a pipe, each molecule is oscillating back and forth parallel to the axis of the pipe. For a given molecule, its amplitude is its maximum displacement from its mean position. The molecules with the greatest amplitude will be at the antinodes (unless I've confused nodes and antinodes again).

In general, we can write a traveling wave as y=A sin(kx-ωt), for suitable choices of x=0 and t=0. y here is the displacement of the particle at position x at time t. The velocity of that particle is therefore Aω cos(kx-ωt), and its peak speed is |Aω|. So its peak KE is as (Aω)². For a given frequency, the energy content is proportional to the square of the amplitude.

It is also interesting to consider interference. Where waves cancel the amplitude is zero, and where they maximally reinforce the amplitudes add up. If the energy were only proportional to the amplitude we would find that the total energy over the area must be less than the sum of the energies of the individual waves. But because it is proportional to the square of the amplitude it turns out that the integral over the area gives the same energy as by adding the energies of the two waves.

 

[randyrandy]

Yes, that's what wave amplitude means. It is the max displacement from the mean position.
For a surface wave on water, the mean position is where the surface would be if not for the wave, so the amplitude is the height of a wave crest above that.
For a longitudinal wave, such as an air column in a pipe, each molecule is oscillating back and forth parallel to the axis of the pipe. For a given molecule, its amplitude is its maximum displacement from its mean position. The molecules with the greatest amplitude will be at the antinodes (unless I've confused nodes and antinodes again).

In general, we can write a traveling wave as y=A sin(kx-ωt), for suitable choices of x=0 and t=0. y here is the displacement of the particle at position x at time t. The velocity of that particle is therefore Aω cos(kx-ωt), and its peak speed is |Aω|. So its peak KE is as (Aω)². For a given frequency, the energy content is proportional to the square of the amplitude.

It is also interesting to consider interference. Where waves cancel the amplitude is zero, and where they maximally reinforce the amplitudes add up. If the energy were only proportional to the amplitude we would find that the total energy over the area must be less than the sum of the energies of the individual waves. But because it is proportional to the square of the amplitude it turns out that the integral over the area gives the same energy as by adding the energies of the two waves.

Thank you for that explanation.

formally speaking, what energy equation would be used on the other side of 1/2kx^2 = ? that uses radius in it's equation?

why are water waves considered harmonic motion? i thought harmonic meant things like springs and pendulums and stuff?

 

[Cutter Ketch]

formally speaking, what energy equation would be used on the other side of 1/2kx^2 = ? that uses radius in it's equation?

I think you have the ideas. You just haven’t put them together. You were thinking correctly when you decided the energy should remain constant and so the energy density had to drop with the circumference. That’s what relates to the radius. Energy conservation. Your only mistake in that reasoning was equating energy to amplitude.

 

[randyrandy]

I think you have the ideas. You just haven’t put them together. You were thinking correctly when you decided the energy should remain constant and so the energy density had to drop with the circumference. That’s what relates to the radius. Energy conservation. Your only mistake in that reasoning was equating energy to amplitude.

I'm still uneasy because I still don't see a concrete formula(s) connecting energy and radius to amplitude... I'm curious to find out what the formal way of solving this would be using nice neat physics formula.