Oscillatory motion and transverse wave on string

  • #1
I'm having some trouble with oscillatory motion...


"A transverse wave on a string is described by the equation
y(x, t) = (0.350 m)sin [(1.25 rad/m)x + (99.6 rad/s)t]
Consider the element of the string at x = 0. (a) What is the time interval between the first two instants when this element has a position of y = 0.175 m? (b) What distance does the wave travel during this time interval?"

ok, the time interval between the first two instants when the element has the same position, that would be the period / 2 right?

so, T = 2Pi/omega omega can be read from the equation and it is 99.6rad/s

so T/2 is what i'm looking for and it is = 0.0315

but that's wrong...the closest answer (on a multiple choice 'quiz') is 31.0 miliseconds, but their answer is exact and so is mine, they are not the same

ok, so maybe i'll use 31ms for part b and see what happens...

well, the distance that the wave would travel would be the angular frequency divided by the wave number k, and all of it multiplied by the period, that would give me the correct units

since omega/k = velocity and velocity * time = distance

the problem is, my answer comes out close to 2.5m and the lowest answers on the multiple choice were 1.68, 1.86, and 3.25, so i'm horribly off but i don't understand why

hints would be appreciated

thanks for your time
 

Answers and Replies

  • #2
3,042
15
is .0210seconds an option? ie 21 miliseconds?
 
  • #3
i believe it was
 
  • #4
3,042
15
Try doing this, set x=0, and see what the resulting equation looks like. Now think about what half the period would resemble on the wave, does this anwser what the question is addressing?



"Now think about what half the period would resemble on the wave,"

I realized I just told you that, but its not very accurate. When you put x=0, the equation you get does not describe the motion of the wave, rather the sin(x=0,t)-'graph', describes the up-down motion at the point x=0, sorry about telling you the wrong thing. But still, think about what half the period would mean.
 
Last edited:
  • #5
hmm, ok, if x is 0, then the equation becomes .175 = .35 sin (99.6t)

how can i visualize this? half of the period would be the next point when x = 0

if i solve for t then

.175/.35 = sin (99.6t) 0.5 = sin (99.6t) so the value of t when x is 0 and y = .175 is (arcsin 0.5 )/99.6 = t that comes out to .0053s

hmm...lol

hmm, this problem shouldn't cause me this much trouble, am i overlooking something obvious?

why doesn't just taking half the period work? that's what the interval between the two closest instances of the wave reaching the same point is...right? so why doesn't 2Pi/omega work?
 
  • #6
3,042
15
hmm, ok, if x is 0, then the equation becomes .175 = .35 sin (99.6t)

how can i visualize this? half of the period would be the next point when x = 0

Exactly, and thats not what the question asked.

You are close now. Try to graph it and see when it intersects the line y=.175. You found one of the points. But the question asked for the time between reaching it and reaching it again, so obviosly you need to solve for two such instances, and find the time between those events. Taking half the period wont for a simple geometric argument. Draw a sin wave on a piece of paper of amplitude 1. If I ask you the time between zero, in that case it IS half the period. But what If I ask you time between say .999. Its def. NOT anywhere near half the period. Draw it and you will see what I mean.
 
  • #7
mezarashi
Homework Helper
651
0
This is how I would have approached the problem. I don't know if these answers will match those that you have:

We know that transverse wave functions can be written:

[tex]y(x,t) = Asin(\frac{2\pi}{\lambda}x + \omega t)[/tex]

Comparing this with your equation, the first question would be asking the same thing as, what is the period of the wave. As [tex]\omega = \frac{2\pi}{T}[/tex] we can directly find it. Same goes for the second question. The second question thus asks, in the time equal to one period, how far has the wave traveled. By definition, it would have travaled one wavelength. From comparing equations again, we see [tex]\beta = \frac{2\pi}{\lambda}[/tex] can be used to solve directly.

I hope that may be useful.
 
  • #8
3,042
15
mezarashi, I dont see how you are doing that. Just graph the equation,

y=.35sin(99.6t), and find the time between the first two instances it crosses .175, it will not equal 2*pi/omega. Help me out.

The first question cant ask the period of the wave, its dealing with only a single point on the wave.
 
Last edited:
  • #9
mezarashi
Homework Helper
651
0
Ah my bad. 0.175 isn't the maximum value. Seems like it is half, or sin 30, sin 150, according to sin (2pit + pi/2) = 0.5. It's turning out to be more trigonometry than physics :P

Then using the change in phase proportion to 2pi radians, we can find the proportion of length compared to the wavelength. That is

[tex]\frac{\Delta \phi}{2\pi} = \frac{\Delta x}{\lambda}[/tex]

Now I hope I'm right this time ;)
 
  • #10
3,042
15
no, you cant have a delta x, were dealing with the time variable here, delta t's.
 
  • #11
mezarashi
Homework Helper
651
0
cyrusabdollahi said:
no, you cant have a delta x, were dealing with the time variable here, delta t's.

Umm, that was for the second part.

I thought the sin30, sin 150 I mentioned should have solved the 'time' problem. We have a case where [tex]\omega t_1 = \frac{\pi}{3}, \omega t_2 = \frac{5\pi}{6}[/tex]
 
  • #12
3,042
15
oh, then im sorry. I thought you were doing the first part still. The second part is easy once you know the first. You know the wave speed, and the amount of time from part a, just multiply the two and your done.
 
  • #13
thanks for the help guys, i got it!
 
  • #14
6
0
Hi, I need to solve this problem, but I do not understand how the time interval is obtained. I tried to follow this thread but do can not understand how the solution is obtained. Do I solve for t, and then add a wavelength to the value to find the next instant of time when the element has a position of 0.175m? Then do I find the difference of the two time intervals obtained?
 
  • #15
1
0
Hello
I have a problem with oscillatory motion detection.
The problem is something like this: I have a data set of the motion of particle, now out of that i need to check whether it is oscillating or not what should I do.
 

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