Calculating Displacement of a Point on a Spring with a Transverse Wave

[ghostbuster25]

Hi guys,

Having trouble with a question

I need to know the displacement of a point on a spring at certain times.

At t=0 point p is at its maximum displacement at 2cm
point A is further down the spring at p-A = 0.4m

A transverse wave of frequency 5Hz is moving along the spring at 1m/s

What is the displacement of A at t=0, t=0.2s, t=0.25s and t=0.3s

I tried using -Acos(frequency)t but I am pretty sure that's not right as i don't know the amplitude. I first thought i could use the displacement along the x-axis but I am pretty sure that's not right.

Any suggestions?

 

[The CdePster]

Heya m8! Firstly in the equation you are using I assume by frequency you mean angular frequency?? It is the right equation though. Also I believe that if you use v=f*wavelength and rearrange to get the wavelength you get a stunning answer which is a 'bizarrely' a multiple of the distance that point A is away from point p. Which makes the whole thing very easy i think because you can easily find the amplitude. I've probably made some fundamental error ha! ;)

 

[ghostbuster25]

i got 0.2m for the wavelength but am not sure how to use this in the equation

would it be -0.2cos5t ?

 

[The CdePster]

Nope, i believe that the equation for the displacement is x=x0 cos(omega*t) where omega(the greek symbol that looks like a pair of melons ;)) is the angular frequency and the angular frequency can be found by doing this omega= 2pi*f. And x0 is the amplitude, not the wavelength.
Draw a cosine wave for the displacement and i think youll find that if you move along the wave two wavelegths(aka 40cm like in the question) so across two peaks of the wave then two wavelengths along is just the same place on the wave as the original place just shifted along by 40cm! Crazy huh? So...at that point assuming that there are no energy losses what would remain the same? If you can work that question out then you can just plug in the numbers! )

 

[ghostbuster25]

hi thanks that kinda makes sense, I am just having trouble with the amplitude...x0...how do i work this out and fit it in?

 

[The CdePster]

It gives you the amplitude in the question lol! :) It say the maximum displacement(aka the amplitude) at time t=0 is 2cm. Hence the amplitude 2 wavelengths down is the same. The wavelength is the distance from peak to peak or from one point in the same place to another on the wave. do you get the whole angular frequency thing, because you have to convert otherwise you will get the wrong answer. that's quite difficult to visualise but its always angular frequency with sine and cosine curves.