# Homework Help: Physics particle on a wave problem

1. Mar 24, 2008

### rnp

Hi, I searched thread similar to my problem but none of the responses helped me to calculate the right answer. I am given the speed (v=400 m/s), wavelength (.16 m) and amplitude (1.4 mm). I need to calculate time for particle of the string to move through a total distance of 1 km. I calculated the period and also how many wavelengths were necessary to get to 1 km and then multiplied (as did the others who had problems with this). I don't understand what to do.

2. Mar 24, 2008

### Staff: Mentor

Can you show us some of your numbers? Please write out your calculation and solution, and we'll see if you're missing something.

3. Mar 24, 2008

### rnp

Hi thanks for getting back to me so soon.
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.

4. Mar 24, 2008

### rnp

Do I divide the total distance of 1 km by .0000028 km (two times the amplitude) to find how many cycles are in the total distance? then divide by the frequency (number of cycles per second)?

5. Mar 24, 2008

### Astronuc

Staff Emeritus
Last edited: Mar 24, 2008
6. Mar 24, 2008

### Staff: Mentor

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.

7. Mar 24, 2008

### Staff: Mentor

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

8. Mar 24, 2008

### rnp

it moves the magnitude of the amplitude above and below the equilibrium and returns to it's starting point of zero by the end of the period.

9. Mar 24, 2008

### rnp

it only says "the amplitude of the wave is 1.4 mm. How much time is required for a particle of the string to move through a total distance of 1.0 km?"
i tried using amplitude as in the way i asked before and still got the wrong answer. something is not clicking. sorry!

10. Mar 24, 2008

### Staff: Mentor

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:

$$y = A sin(\omega t)$$

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

11. Apr 15, 2008

### mr_schticker

just to help anyone out that searched for this...

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!