Simple Harmonic Motion Fitzpatrick 3U Chapter 25(C) Q4(a) (1 Viewer)

[Kingportable]

Simple Harmonic Motion Fitzpatrick 3U Chapter 25(C) Q4(a)
The displacement x at time t of a point moving in a straight line is given by x=asin(nt+epsiton). Find the form wich this expression takes if initially:
b) x=0 and the velocity is negative

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Hey i'm trying to this question and currently working on question "a"
a) x'=0 and x=-5
I kind of got that since x'=0 and x=5
using v^2=n^2(a^2-x^2)
n^2(a^2-25)=0
so either n=0
or a^2-25=0
a^2=25
therefore amplitude is a==+/-5
a=-5 is a partial solution since a>0

so form is... x=5sin(nt+epsiton)
but the answer says its x=5sin(nt+(pi/2)
i have no idea how to solve for epsiton
All i know is displacement is x=asin(nt+alpha)
V^2=n^2(a^2-x^2)
x''=-n^2(x)

 

[lolcakes52]

okay, at the initial time t=0. sub that into x=5sin(nt+e) where e is epsilon. solve for e and you get e=pi/2

 

[Kingportable]

Thanks dude, thats right since the this is the sinx function then at t=0, v=0 and v'=0. Just a thought, if i put sin^-1(0) it comes up with zero --> however i do know that that meant x=0, pi/2, pi, 3pi/2 and 2pi or x=0,90,180,270 and 360. How will you choose? Why can't i just say epsiton=0?

 

[Kingportable]

Any thought on question B? I used that the rationalisation that is dx/dt = -(dx/dt) then x=-x.

I use the same idea that at at=0, x=0 and solved for epsiton. I got that epsiton=sin^-1(0)=0
in the first question i looked at the back of the page and it said i need a pi/2 and interpreted the same thing for pi/2
Thats different this time, since it required epsiton=0
so i got the required x=-asin(nt)
however i do not think this is right because for both questions i had to look at the answer first to make a decision on i see sin^-1(0)=epsiton