SHM cambridge question (1 Viewer)

[wandering17]

A particle moving in simple harmonic motion has period pi/2 seconds. Initially the particle
is at x = 3 with velocity v = 16 m/s.
(a) Find x as a function of t in the form x = bsin nt + ccosnt.

So i got x = 4sin4t+3cos4t

But for part b)
How do i show Find x as a function of t in the form x = a cos( nt - e)

 

[Drongoski]

So i got x = 4sin4t+3cos4t

But for part b)
How do i show Find x as a function of t in the form x = a cos( nt - e)

Then one approach is to draw a right-angled triangle with, relative to one angle @, you have adjacent = 3, opp = 4 so that hypotenuse = 5. In this case cos @ = 3/5 and sin @ = 4/5.

.: 3cos 4t + 4sin 4t = 5[(3/5)cos 4t + (4/5)sin 4t] = 5[cos@cos 4t +sin@sin 4t] = 5cos(4t - @)

where @ = inverse sin(4/5)