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berkeman

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v=lambda/period so 400 m/s = (.16 m)/ T; T = .0004 seconds

frequency= 1/T so this is 2500 cycles per second. 1 cycle is equal to the wavelength of .16 m. The total distance of 1 km/.00016 km shows there are 6250 wavelengths in 1 km. multiplying this by the time it takes to complete one wavelength i got 6250 X .0004 = 2.5 seconds. I don't see how the amplitude ties in to my calculations.

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Astronuc

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OK, try to think of the relationship between wave velocity, wavelength and frequency. Here is a good reference.

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html#c2

Now in a transverse wave, in one wavelength, how far does a particle move (hint: amplitude) in one period or one wavelength?

http://hyperphysics.phy-astr.gsu.edu/hbase/wavrel.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html#c2

Now in a transverse wave, in one wavelength, how far does a particle move (hint: amplitude) in one period or one wavelength?

http://hyperphysics.phy-astr.gsu.edu/hbase/wavrel.html#c1

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berkeman

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I thought that might be what was tripping you up. I interpret this question to be how long does it take a particle at a fixed point (like on a string) to move up and down a total of 1km, not move along with the wave 1km. So the amplitude is the peak-to-peak distance that the particle moves as the wave goes by. if the p-p amplitude is 1.4mm, then the particle moves up 1.4mm and back down 1.4mm as one wavelength goes by. Read the question closely, however, because the 1.4mm may be the peak amplitude, or may be the peak-to-peak amplitude. There's a 2x difference.

v=lambda/period so 400 m/s = (.16 m)/ T; T = .0004 seconds

frequency= 1/T so this is 2500 cycles per second. 1 cycle is equal to the wavelength of .16 m. The total distance of 1 km/.00016 km shows there are 6250 wavelengths in 1 km. multiplying this by the time it takes to complete one wavelength i got 6250 X .0004 = 2.5 seconds. I don't see how the amplitude ties in to my calculations.

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berkeman

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Ack, Astro beats me out again! When will it stop!? :-)

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i tried using amplitude as in the way i asked before and still got the wrong answer. something is not clicking. sorry!

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berkeman

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Well, you correctly got the frequency as 2500Hz, so 2500 times per second, the particle will go from zero displacement, up to max displacement, down to maximum negative displacement, and back to zero displacement. That total motion is either 2x 1.4mm, or 4x 1.4mm, depending on whether the "amplitude" they are giving you is the peak-to-peak amplitude, or the peak amplitude. My guess is it is the peak amplitude, since the multiplier in front of the sin() function is often referred to as the amplitude:

i tried using amplitude as in the way i asked before and still got the wrong answer. something is not clicking. sorry!

[tex]y = A sin(\omega t)[/tex]

(checks for a reply by Astro before hitting Submit Reply.....)

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It makes sense now how you handled it, but I still had some problems conceptually finishing it. But thanks so so much, I'd probably never have solved this without the kickstart!

To help tie up the loose ends, now that I solved it, here's how it works:

follow the process as above to get the frequency (cycles per second) and then you have to multiply the amplitude (in meters) by 4, as suggested above. divide the 1000 m (1 km) by your new total for amplitude (amplitude x 4) and then divide that by your frequency, because that is the total time it will take for your amplitude to equal a km.

And thats it!