What expressions do you have for the relationship between amplitude and energy or power?randyrandy said: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?
Think about a single circular wavefront. As it travels out it is distributed over a greater...?randyrandy said:about energy distributed over a greater area which translates to amplitude?
Not exponentially.randyrandy said:something drops off exponentially as you go further out
Then that answer is wrong!randyrandy said:But the answer says it's 1cm not 0.5cm...
randyrandy said:But the answer says it's 1cm not 0.5cm...
You overlooked this part of my post:randyrandy said:But the answer says it's 1cm not 0.5cm...
Or think about SHM, which is essentially wave motion. How does energy in a spring relate to displacement?haruspex said:The way amplitude relates to energy is crucial. What do you know about waves?
HallsofIvy said:Then that answer is wrong!
Then I will ask you againrandyrandy said:I'm confused more than ever...
haruspex said:How does energy in a spring relate to displacement?
1/2kx^2? = energy. but that formula doesn't have radius as a variable.haruspex said:Then I will ask you again
Cutter Ketch said:No, I’m afraid it is right.
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 said:1/2kx^2? = energy. but that formula doesn't have radius as a variable.
haruspex said: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.
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 said:I found this formula for a spring F=kx. Would that be applicable here?
haruspex said: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.
Yes, that's what wave amplitude means. It is the max displacement from the mean position.randyrandy said:Would the problem assume that 2cm is the max displacement at 10cm from center?
Thank you for that explanation.haruspex said: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 said:formally speaking, what energy equation would be used on the other side of 1/2kx^2 = ? that uses radius in it's equation?
Cutter Ketch said: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.
The amplitude of a circular wavefront is the maximum displacement of the wave from its equilibrium position. It represents the strength or intensity of the wave.
Yes, the amplitude of a circular wavefront can change depending on the source of the wave and the medium it travels through. It can also change due to factors such as interference and diffraction.
The amplitude of a circular wavefront can be measured by finding the maximum displacement of the wave from its equilibrium position. This can be done using instruments such as an oscilloscope or by analyzing the wave equation.
The amplitude and energy of a circular wavefront are directly proportional. This means that as the amplitude increases, so does the energy of the wave. This is because a larger amplitude represents a stronger wave with greater energy.
The amplitude of a circular wavefront does not directly affect its wavelength or frequency. However, it is indirectly related as a higher amplitude wave may have a shorter wavelength and higher frequency compared to a lower amplitude wave. This is because the energy of the wave is distributed over a smaller distance, resulting in a higher frequency.