Calculus Applied to Simple Harmonic Motion

Guide 13-2. Calculus Applied to Simple Harmonic Motion (optional)

Physics or Math?

While the derivations below show how to use calculus to obtain the relationships v_max = Aω and a_max = Aω², any good AP Physics B student must be able to obtain these relationships using physics. The first relationship is obtained using a conservation of energy argument, and the latter relationship is obtained using a net force argument.

The equation for period, T = 2π(m/k)^1/2, requires calculus for its derivation. Therefore, this equation is taken as given for problems in this class.

In Calculus, you've probably learned already how to take derivatives of sine/cosine functions and how to use the chain rule. Here's an application to the subject of oscillations. We'll start with the equation describing the position of an object oscillating under simple harmonic motion: x = Acos(ωt). Now we could just as well have started with x = Asin(ωt). One difference is that for the cosine function, the initial position is A, and for the sine function, the initial position is 0. Other differences have to do with the initial velocities and accelerations. In the former case, the initial velocity is 0 and the initial acceleration is maximum. In the latter, the initial velocity is maximum and the initial acceleration is zero. Basically, the position, velocity, and acceleration functions are shifted by 90° or a quarter cycle.

Let's see how to quickly get the velocity from the position. Velocity is the time derivative of position. In the following derivation, we use the fact that the derivative of the cosine is the negative of the sine. We also use the chain rule in taking the derivative of the argument, ωt, of the cosine. Sinceω is constant, the derivative of ωt with respect to time is simply ω.

Examining the final result, note that v_max = Aω. This is consistent with the result in the textbook. Note also that the velocity is shifted 90° from the position. This is something you found in M07t.

We can repeat the above process to find the acceleration. Just take the derivative of velocity with respect to time.

Again, the result a_max = Aω² is consistent with that found in the text. The negative sign in front of the cosine function means the acceleration is shifted 180° from the position.

Since a = -Aω²cos(ωt) and x = Acos(ωt), then by substitution, a = -ω²x. We also know from a net force analysis that a = (-k/m)x. By comparison of the two expressions for a, we see that ω² = k/m, which, after substituting ω = 2π/T and doing some algebra leads to T = 2π(m/k)^1/2.