# Thread: oscillation with trig formula

1. ## oscillation with trig formula

An oscillating particle has displacement y metres, where y is given by y=asin(kt + A)°, where a measured in metres, t is measured in seconds and k and A are constants. The time for a complete oscillation is T seconds.
Find k in terms of T.

so T must be 4x(length of time taken for the particle to move from y=0 to y=a its maximum displacement)
This length of time therefore can be found using sin(kt + A)°=1
so t=(90 - A)/k
so
T=4t=(360-4A)/k

now to eliminate A
y=0 at t=0, meaning 0=asin(0 + A)
so sinA = 0, A=0

T=360/k

Is this reasoning correct?

2. ## Re: oscillation with trig formula

I think you can write also (a shorter way):
In general if you've an oscillation given by:
$y=A\sin(\omega\cdot t+a)$
Then the pulsation $\omega$ is $\omega=\frac{2\pi}{T}$ where T is the period. In this case $\omega=k$ therefore $k=\frac{2\pi}{T}\Leftrightarrow T=\frac{2\pi}{k}$

3. ## Re: oscillation with trig formula

sorry bro no idea what a pulsation is, never learnt, so i'm pretty sure i can't use it in my answer
thanks though i'll think of you when i get around to learning it