Simple Harmonic question! (1 Viewer)

[FellowStudent]

A particle moves with simple harmonic motion. If the particle starts from the equilibrium position (centre of motion) with velocity 4m/s toward the origin and the period is pi/2 seconds, find:
a) the displacement at time t
b) the amplitude

Thank you!

 

[Sp3ctre]

Imo the question isn't very well worded as it's hard to distinguish whether v is meant to be positive or negative, but I'll guess that they mean v=-4 in this case.

a) a = -n^2x
given that T=π/2, then π/2 = 2π/n (according to the equation T = 2π/n)
nπ = 4π
n = 4
∴ a = -16x
d[(1/2)v^2)]/dx = -16x
(1/2)v^2 = ∫-16x dx
(1/2)v^2 = -16x^2/2 + c
v^2 = -16x^2 + c
when x=0, v=-4
∴ c=16
v^2 = -16x^2 + 16
v^2 = 16(-x^2 + 1)
v = 4sqrt(1 - x^2)
dx/dt = 4sqrt(1 - x^2)
dt/dx = 1/4sqrt(1 - x^2)
t = 1/4∫1/sqrt(1 - x^2) dx
t = 1/4 sin^-1(x) + c
when t=0, x=0, ∴ c=0
∴ t = 1/4sin^-1(x)
4t = sin^1(x)
x = sin(4t)
∴ at time t, x = sin(4t)

b) Since the general form of a particle undergoing SHM is x = Acos(nt+θ),
from the eq'n x = sin(4t), A=1