Projection of a uniform circular simple harmonic motion
From the right triangle highlighted in the figure above, comes:
As a p = ω 2 .R, R = A.θ, θ = θ 0 + ω.t, it follows:
a(t)= -ω 2 .A.cos(θ 0 +ω.t)
Time function of the MHS acceleration
The sign (-) in the above equation is necessary because, at the instant in question, the acceleration α has a direction opposite to that of the Ox axis.
We can conclude that:
– the magnitudes of the acceleration α and the elongation x (from the hourly elongation function) are directly proportional. We can demonstrate this proportion as follows: just substitute the time function of elongation in the time function of acceleration: α(t) = -ω 2 .A.cos(ω.t + θ 0 ) ex(t) = A.cos( ω.t + θ 0 ). In this way we arrive at the following equation:
α = -ω 2 .x
We can also demonstrate that the period T of the mass-spring oscillator, which describes an MHS, depends only on the mass m and the spring constant k of the spring, so it does not depend on the amplitude of oscillation of the MHS. This demonstration starts from the expression of Newton’s 2nd Law :
F R = m . a
Since a= α = -ω 2 .x, , and m being the mass of the oscillating body, we have:
F R = m .(-ω 2 .x) ⇒ F R = -(m .ω 2 ).x
As F R = -k .x, then k = m.ω 2 , from which:
