oscillation with trig formula

[red and white kop!]

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?

 

[Deleted member 4993]

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?

First of all - conventional way to write that equation is:

y = A * sin(w*t + B).........do not write "asin" because lot of time that is shorthand for "arcsin"

Another point, use radians as opposed to degrees.
Otherwise your work is correct. So your answer would be:

T = 2*pi/k

 

[red and white kop!]

my textbook used degrees in this section.
sorry about the text, when i typed this in on microsoft word (i posted this on several sites) i used italics for the variables, i forgot they didnt work here.
cheers, thanks for your time

 

[wjm11]

Find k in terms of T.

Just rearrange your answer:

k = 360/T

or

k = 2pi/T