Simple Harmonic Motion Fitzpatrick 3U Chapter 25(C) Q7(a)

[Kingportable]

This question's answer was completely different to mine.

Question 7 (CHAPTER 25C)
Solve the differential equation d^2x/dt^2 + 16x = 0 subject to the conditions x=3 and dx/dt=16 when t=0. Find the maximum displacement and the maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.


what i did
i said that x=asin(nt+alpha)
and that d^2/dt^2 +16x=0
is the same as x''+16x=0
so x''=-16x <-- SHM where n=4

Since this is a sin function at t=0, x=0
so asin(alpha)=0
alpha=Sin^-1(0)
so, alpha=0 at t=0

since v=16 at t=0
i got that v=ancos(nt+alpha)=16
since at t=0, alpha =0
ancos(0)=16
an=16
since n=4
4a=16
a=4
so... x=4sin(4t)
however the back of John Fitzpatrick 3U says
x=4sin(4t) +3cos4t

the 4sin(4t) part makes total sense.... BUT WHERE THE HELL DID THAT 3cos4t COME FROM!!!!!!!!....WTTTFFFFF??????
So yeah im stuck on finding the displacement, is the axiliary methods in play in here?

Anyways the other answers are: 5m;20m/s

[Mr Slick]

umm acceleration = -16x
use d/dx (v^2 / 2) to find velocity

then sub x=0 to find max speed

then put v=0 to find x i.e. amplitude

[Mr Slick]

yep i am right...i get the correct answer!!!!!!!

[My Name is Khan]

+1

umm acceleration = -16x
use d/dx (v^2 / 2) to find velocity

then sub x=0 to find max speed

then put v=0 to find x i.e. amplitude

[My Name is Khan]

We can use the other method to solve this question as stated by Mr Slick

[Kingportable]

x=4sin(4t) +3cos4t

Ok thats all good, but where did this come from!

[zb123]

It's not a a sine function as when t=0, x=3, you said x=0 when t=0, but clearly the question says different.