help a girl out please :(

[brookelynne3]

:cry:Ok I don't know exactally what I am doing wrong here, someone please help!:cry:

The speed of a transverse wave on a string is 400 m/s, and the wavelength is 0.19 m. The amplitude of the wave is 1 mm. How much time is required for a particle of the string to move through a total distance of 1.0 km?


I tried to solve for this by:
v=lamda/T so..
400=.19/t which equals a time of
.000475 seconds per cycle and i thought if I want 1000m (1km) i need to go 1000m/.19m to get how many cycles are necessary, that equals 5263.15 cycles to go 1000 meters so...
5263.15 cycles (.000475m/cycle)=2.5 seconds to complete all the cycles that equals 1000 meters or 1 km.
Except when I input this into my computer, its saying its wrong. can anyone pelase tell me what I am missing???

[marcusl]

No, a particle on the string moves transversely only. It never moves down the string along the .19 m wavelength.

Each cycle, the particle moves 1 mm in each direction, for a 2 mm total motion. You can take it from there to find the number of cycles, and time, to go 1 km.

[brookelynne3]

I dont get what to do with that, I really cant take it from there..
do Imultiply the amount of cycles by .002m and then by time or do I use the .002m to find the amount of cycles?

[berkeman]

marcusl is pointing out that your problem statement is incorrect. The particles of the string only move up and down with the wave rhythm, not down the length of the string (nothing moves 1km down the string except the waves). Could you please try posting the exact text of the question?

[andrevdh]

For each cycle it will cover a distance of four millimeters. That is two down and two up. One needs to be careful in calculating the elapsed time for the fractional cycles though, since it is not motion at a constant speed.

[belliott4488]

Your approach was correct, except that when you figured out how many cycles it would take to total 1000m, you used the wrong distance for the distance traveled per cycle. Use instead the distance suggested in the other posts, in the same calculations you did before, and you should be fine.