physics wave help: period vs frequency

[dmission]

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when you have just a standard graph of a wave (displacement v. time), do you visually see the period or the frequency? it's the period right?
e.g, longer wavelength = longer period?

 

[PingPongPro]

Frequency = # of cycles per second. (Hz)
Period= Time it takes to complete 1 cycle.

You can get both of these values from a standard graph.

 

[dmission]

Frequency = # of cycles per second. (Hz)
Period= Time it takes to complete 1 cycle.

You can get both of these values from a standard graph.

Thanks, but what about my question? 😉

 

[Atin2011]

well you see both. Frequency(cycles/sec) and Time period(sec) have inverse relationship to each other in the wave function. so you should be able to calculate one from other.

 

[dmission]

well you see both. Frequency(cycles/sec) and Time period(sec) have inverse relationship to each other in the wave function. so you should be able to calculate one from other.

Ok, bigger wavelength means bigger period right?

 

[PingPongPro]

Well, what you "see" is a line. There is no clear cut answer, since I don't know what you are looking at...

Its like looking at kinematics graph of distance vs time. You "see" the velocity and you "see" the distance. If you try put on your glasses, you can probably "see" the acceleration if there is one. Likewise, I don't know what you are looking at, so I cannot say for sure. Maybe post a picture with arrows pointing and it would be more clear.

With my definitions i posted and a graph, it shouldn't be too difficult to get the period and frequency.

 

[dmission]

Well, what you "see" is a line. There is no clear cut answer, since I don't know what you are looking at...

Its like looking at kinematics graph of distance vs time. You "see" the velocity and you "see" the distance. If you try put on your glasses, you can probably "see" the acceleration if there is one. Likewise, I don't know what you are looking at, so I cannot say for sure. Maybe post a picture with arrows pointing and it would be more clear.

With my definitions i posted and a graph, it shouldn't be too difficult to get the period and frequency.

Thanks; Mostly just wondering now if I'm correct in assuming larger period would mean a larger wavelength and vice-versa?

 

[indianjatt]

For a displacement versus time graph, the distance between trough to trough or crest to crest will be the period. From the period, you can get the frequency by getting the inverse. I hope that answers your question.

 

[dmission]

For a displacement versus time graph, the distance between trough to trough or crest to crest will be the period. From the period, you can get the frequency by getting the inverse. I hope that answers your question.

Yep. Thanks. So longer period means longer wavelength, and vice-versa.

 

[indianjatt]

Yep. Thanks. So longer period means longer wavelength, and vice-versa.

Correct.

 

[whiteshadodw]

helps to understand the correct units for these

f = cycles/sec
T = sec/cycle

 

[minombre]

Period and frequency have to do with a wave's oscillations in time (the x axis is time), while wavelength has to do with a wave's oscillations in space (the x axis is distance). Without knowing the wave's velocity, there's no way to know the relationship b/w period and wavelength.

If you know the velocity, you can use:

v=(f)(lambda) = (lambda)/(T).

It's easy to see that if v~1, then wavelength~T; if v<1, then T>wavelength; and if v>1, wavelength>T.

 

[BerkReviewTeach]

when you have just a standard graph of a wave (displacement v. time), do you visually see the period or the frequency? it's the period right?
e.g, longer wavelength = longer period?

You can read the period directly from the graph you described. If the amplitude starts at 0 then goes (+) then back to 0 then goes (-) and finally back to 0, that represents one cycle (one wavelength), then the time it takes for that cycle is the period. The frequency is simply the inverse of the period, so you can get the freqeuncy from that graph too.

However, from the graph you described, there is no way to conclude anything definitive about the wavelength. The distance you gave was for displacement from equilibrium (i.e., amplitude). The graph does not give us any information about propagation distance, so we cannot conclude anything about propagation distance per time (the wavelength).

Correct.

Actually, you can't reach a conclusion without additional information. Knowing only the frequency of a wave does not tell us enough information to conclude anything about the wavelength. Consider red light and blue light in a vacuum. Both have the same speed, so a lower frequency (larger period) corresponds to a longer wavelength (so it would be a valid conclusion in that case). Now consider red light in a vacuum versus red light in air. Both have the same frequency, but the slower wave (the one in air) has a shorter wavelength than the faster wave (the one in a vacuum). The conclusion doesn't hold in this case.

There are too many other variables to reach any conclusion besides the period (and frequency) from the standard displacment vs. time graph described by the OP.