2014 HSC Maths Ext multiple choice question HELP! (1 Viewer)

[narges-ts]

OK so obviously for multiple choice they give you the answers but the problem is the don't show you how to get that answer. So the question I was stuck on was Q.7:

A particle is moving in simple harmonic motion with period 6 and amplitude 5.
Which is a possible expression for the velocity, v, of the particle?

ANS= 5pi/3 cos (pi/3)t

 

[jazz519]

The basic equation used is x=Acos(nt+a)

Period = 2pi/n

Thus, n=Pi/3

A=5

Let t=0, when x=0

0=5cos(a)
a=Pi/2

Thus, x=5cos(t*Pi/3+Pi/2)
Differentiate for the velocity

V=-5pi/3sin(t*pi/3+Pi/2)
V=-5pi/3cos(-t*pi/3) since sinx=cos(90-x)
V=-5pi/3(t*pi/3)

Direction I don't think is that important cause you could just define one way as positive or negative

 

[narges-ts]

if i differentiate x=5cos(Pi/3)t wouldnt the cos turn into -sin??

 

[jazz519]

if i differentiate x=5cos(Pi/3)t wouldnt the cos turn into -sin??

I fixed it up