Mechanics

# Time function of acceleration in MHS

Projection of a uniform circular simple harmonic motion

The acceleration α of the point Q , describing the MHS, is obtained by projecting the centripetal acceleration p , of the point P describing the MCU, on the axis Ox.

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 :

R = m . a

Since a= α = -ω 2   .x, , and m being the mass of the oscillating body, we have:

= m .(-ω 2   .x) ⇒ F  = -(m .ω 2 ).x

As F  = -k .x, then k = m.ω 2 , from which: