I am currently self-studying introductory physics and have come across an anomaly while attempting to derive the equation for the period of simple harmonic oscillations on a spring. This is how I did it; let me know where I went wrong.

Suppose there is a block of mass m at rest on a frictionless horizontal surface. Onto the side of this block is attached a spring (which obeys Hooke's law) that is also attached to a fixed wall. The spring is compressed a distance x and released, leading to SHM.

At the moment that the spring is compressed the full distance x, the magnitude of the force on the block is given by F = kx. From the time that the block is fully compressed to the time that it recovers to the equilibrium position, the average force exerted on it by the spring is 1/2kx.

Via Newton's second law, I calculated the average acceleration of the block during this same time frame, a = kx/2m.

Then I used kinematics to find the time it took for the block to complete this journey:

x-xo = vxot + 1/2at^2, x = 1/2(kx/2m)t^2, t^2 = sqrt(4m/k), t = 2sqrt(m/k)

Since this time frame is one-fourth of the total period, the period of the motion would be 8sqrt(m/k).

I know the real period is 2πsqrt(m/k), so what went wrong?

I know calculus but would prefer an algebra-based explanation.

• Kinematics only works when acceleration is constant. If you set up a differential equation for position vs. time, and integrate it, you will end up with a sinusoidal functional form. The resulting equation can be used to solve for period. Commented Jul 8, 2018 at 18:26

The error in your derivation is that you are averaging over distance, but then treating it as an average over time; the object undergoing SHM does not spend equal amounts of time at each location.

Strictly algebraically, SHM must be written in terms of cosine (or sine if you like), then it is simple to exploit the period of those functions to find the period of oscillation

This is a good idea, but basically you need to learn calculus.

From the time that the block is fully compressed to the time that it recovers to the equilibrium position, the average force exerted on it by the spring is 1/2kx.

This is not right because the block moves slowly when the spring is more stretched out and quickly when the spring is short. There is more time with higher forces then with lower forces, so the average force is more than $\frac12 kx$

Then I used kinematics to find the time it took for the block to complete this journey: x-xo = vxot + 1/2at^2

That's an equation for constant acceleration. Taking the average acceleration and plugging it in there doesn't work.

In other words, you'll need to learn how to deal with time-varying quantities before you will be able to pull derivations like this off. That's what calculus is for.

• I know calculus. Could you explain it with calc then? Commented Jul 8, 2018 at 19:16
• You may know calculus, but do you know how to solve differential equations like $m \dfrac{d^2 x}{dt^2} + kx = 0$ with initial conditions like $x = A$ and $\dfrac{dx}{dt} = 0$ when $t = 0$? Commented Jul 8, 2018 at 19:34