Answer:
amplitude A = 5
vertical shift = 0
period T = 10 s
frequency = 1/T = 0.1 Hz
angular frequency ω = π/5
h(t) = 5sin(πt/5)
Step-by-step explanation:
An oscillator cannot move between ± 5 units, be at rest at t=0 and at maximum 2.5 s later without some serious external acceleration being applied. It just does not work.
I will ASSUME that the particle is at maximum velocity at t = 0 and use the remaining inputs as stated
As h(0) = 0 the Sine function fits nicely as sin0 = 0
A = 5 as the particle swings between +5 and -5
h(t) = 5sin(ωt)
ω will have units of rad/s
We know the particle will move from max to min position in half the period.
so the period T = 2(7.5 - 2.5) = 10 s
There are 2π radians in each cycle, therefore
ω = 2π rad/10 s = π/5 rad/s
h(t) = 5sin(πt/5)
