Smilebuffalo
Member
3.c) A particle moves in a straight line and its position at time t is given by:
x=1+sin4t+ (root3)cos4t
(i) Prove that the particle is undergoing simple harmonic motion about
x = 1.
When i was reading the exam report, they said it was incorrect to do the following:
a = −16 sin 4t − 16(root3)cos 4t
= −16(sin 4t + (root3)cos4t)
= −16(x − 1) .
This got me confused. I assumed we were always meant to show a = -n^2(x-b)
can someone explain?
x=1+sin4t+ (root3)cos4t
(i) Prove that the particle is undergoing simple harmonic motion about
x = 1.
When i was reading the exam report, they said it was incorrect to do the following:
a = −16 sin 4t − 16(root3)cos 4t
= −16(sin 4t + (root3)cos4t)
= −16(x − 1) .
This got me confused. I assumed we were always meant to show a = -n^2(x-b)
can someone explain?