# Time function of acceleration in MHS

Projection of a uniform circular simple harmonic motion

**α**of the point

**, describing the MHS, is obtained by projecting the centripetal acceleration**

*Q***, of the point**

*a*_{p}**describing the MCU, on the axis Ox.**

*P*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

**and the spring constant**

*m***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**

*k*^{:}

**F _{R} = m . a**

Since *a= α = -ω ^{2} .x,* , and

**being the mass of the oscillating body, we have:**

*m**F _{R } = m .(-ω ^{2} .x) ⇒ F _{R } = -(m .ω ^{2} ).x*

As F _{R } = -k .x, then k = m.ω ^{2} , from which: