Simple Harmonic Motion Question. (1 Viewer)

[SoCal]

Can someone help me out with this question please. If you have the S.B.Jones & K.E.Couchman text book you can see the question on page 183 Q10. If not it reads like this:

A particle travels in simple harmonic motion such that its distance in metres from the centre of motion is given by x = a cos (3t + ) at time t seconds. If the particle starts from rest at a distance of 5 m from the origin, find the distance from the origin after 3 seconds.

Thank you for any help.

 

[Harimau]

Let x= a cos (3t+b), say

Differentiate to get velocity

V= -3a sin (3t+b)

now its given that at t=0, x=5 and v=0

Sub this information into both equations. You get:

5= a cos (b) ..... (1)
0= - 3a sin (b) ..... (2)

Solving simulatiously, you get: a=5 b=0

Therefore, x= 5cos (3t) is your general equation of motion.

Subbing in t=3, you get x= 5cos (9) m

I think its right, and don't forget its in radians...

 

[NSBHSchoolie]

I get the same answer.

 

[SoCal]

That is the right answer but how did you know how to differentiate the first expression x = a cos (3t+b)?

Thank you for helping me out too guys/girls.

 

[LadyMoon]

Because in the equation x=acos(3t+b)
it is known to be a fact that 'a' is a constant so is 'b'

thus when you want to find velocity v, remeber that v=dx/dt
therefore you differentiate x=acos(3t+b) with respect to 't', because it is the only variable!

 

[SoCal]

Originally posted by LadyMoon
Because in the equation x=acos(3t+b)
it is known to be a fact that 'a' is a constant so is 'b'

thus when you want to find velocity v, remeber that v=dx/dt
therefore you differentiate x=acos(3t+b) with respect to 't', because it is the only variable!

I can see it now, thanks heaps LadyMoon. I can't believe I did not see that.