wave energies

[plzNOCarribbean]

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For electromagnetic radiation in general, E = hf is used to calculate energy. Not just for a single photon. Yes, I have looked over the photoelectric effect before. This is another facet of the discussion.

I think I got it.. it seems like for transverse waves and longitudinal waves energy is measured by amplitude, but for transverse waves frequency is used in calculation (i.e. relative energy differences of visible, uv, and infrared E=hf). However, for longitudinal waves like sound waves (amplitude is used to measure energy and intensity, while frequency only measures pitch)...

^So I found one thread from last year to try and clarify this but I am still confused, as the responders didn't really answer the question definitively.

So, based on the prep books, (and wikipedia) a transverse wave is any way that requires a medium to travel; based on this premise, we could eliminate EM radiation as a transverse wave since it can travel in a vacuum (even though it can travel through air, water, etc. It also stated that mechanical waves are transverse, longitudinal, or surface waves.


Based on the above differentiation of wave types , I was wonder, for MCAT purposes, is it a good rule of thumb to assume
(1) that the energy associated with light is proportional to frequency??
(2) and that the energy of transverse waves is proportional to the amplitude?

^is this what we should assume? In EK, they mention that a wave on a rope will have more energy the more you displace it from its equilibrium position (by moving your hand higher when you displace the rope, thus giving this wave more energy). But then again, this wave on a rope has an apparent frequency just like EM waves, and although EM waves are not mechanical waves, they are transverse waves, and their energies are based on frequency; so does frequency dictate/play a role in the way we calculate the energy associated with a transverse wave, like the one on the rope?


really confused about this. I like to think of things as black/white and definitive yes or no's, but I wont be surprised if there are exceptions to this;

 

[costales]

EM waves are actually transverse. Yes, frequency plays a huge role in all wave phenomena - the energy of a light source is always proportional to its frequency.

 

[Majik]

I'll admit this question confuses me too. Sometimes the questions are straightforward, like: "What happens to the energy of an EM wave when frequency increases?" In which case, you'll find that the energy should increase based on the equation: E=hf.

However, for a mechanical wave (a wave that requires a medium to travel), things aren't that straightforward. I'm not entirely sure if we could use the E=hf equation in such a situation (although I have seen the same equation used for sound waves). However, to be safe what I consider is the changes in intensity of the wave and then relate that to energy.

Intensity is proportional to:
- The square root of frequency
- The square root of amplitude
- Velocity of the wave

Let's say you doubled the frequency of a mechanical wave. Doubling the frequency would then Quadruple the Intensity of the wave.

Well, what is intensity then?

Intensity = Power / Area (where Power = Work / time), therefore:
Intensity = Work / (Area x Time); Work ~ Energy
Intensity x Area x Time = Work (Energy)

If the intensity increases by a factor of 4, the energy should also increase by a factor of 4. But you're forgetting one thing: If frequency doubles, the period is cut in half, which means that time is cut in half as well. This would halve the work (energy) done. The overall effect is that energy would double (not quadruple). This is similar to what E=hf would tell us, so technically I think it's still valid to use this equation. However this equation doesn't tell us what happens when things like amplitude or velocity are varied so that's where the intensity equation might be more useful. As a side note, frequency and amplitude are independent of each other while the velocity of the wave depends on the medium the wave is traveling through. Also, the frequency depends on the source that emitted it and remains constant during its propagation.