Students will learn to calculate periods, frequencies, etc. of spring systems in harmonic motion.
Key Equations
\begin{align*} T = \frac{1}{f}\end{align*} ; Period is the inverse of frequency
\begin{align*} T_{\text{spring}} = 2\pi\sqrt{\frac{m}{k}}\end{align*} ; Period of mass m on a spring with constant k
\begin{align*} F_{sp} = kx \end{align*} ; the force of a spring equals the spring constant multiplied by the amount the spring is stretched or compressed from its equilibrium point. The negative sign indicates it is a restoring force (i.e. direction of the force is opposite its displacement from equilibrium position.
\begin{align*} U_{sp} = \frac{1}{2} kx^2 \end{align*} ; the potential energy of a spring is equal to one half times the spring constant times the distance squared that it is stretched or compressed from equilibrium
 In harmonic motion there is always a restorative force, which attempts to restore the oscillating object to its equilibrium position. The restorative force changes during an oscillation and depends on the position of the object. In a spring the force is given by Hooke’s Law: \begin{align*}F = kx\end{align*}
 The period, \begin{align*}T\end{align*}, is the amount of time needed for the harmonic motion to repeat itself, or for the object to go one full cycle. In SHM, \begin{align*}T\end{align*} is the time it takes the object to return to its exact starting point and starting direction.
 The frequency, \begin{align*}f,\end{align*} is the number of cycles an object goes through in \begin{align*}1\end{align*} second. Frequency is measured in Hertz \begin{align*}(Hz)\end{align*}. \begin{align*}1\ Hz = 1\end{align*} cycle per sec.
 The amplitude, \begin{align*}A\end{align*}, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object. The amplitude can vary in harmonic motion, but is constant in SHM.
Example 1
Watch this Explanation
Simulation
Mass & Springs (PhET Simulation)
Time for Practice
 A rope can be considered as a spring with a very high spring constant \begin{align*}k,\end{align*}so high, in fact, that you don’t notice the rope stretch at all before it “pulls back.”
 What is the \begin{align*}k\end{align*} of a rope that stretches by \begin{align*}1\;\mathrm{mm}\end{align*} when a \begin{align*}100\;\mathrm{kg}\end{align*} weight hangs from it?
 If a boy of \begin{align*}50\;\mathrm{kg}\end{align*} hangs from the rope, how far will it stretch?
 If the boy kicks himself up a bit, and then is bouncing up and down ever so slightly, what is his frequency of oscillation? Would he notice this oscillation? If so, how? If not, why not?
 If a \begin{align*}5.0\;\mathrm{kg}\end{align*} mass attached to a spring oscillates 4.0 times every second, what is the spring constant \begin{align*}k\end{align*} of the spring?
 A spider of \begin{align*}0.5\;\mathrm{g}\end{align*} walks to the middle of her web. The web sinks by \begin{align*}1.0\;\mathrm{mm}\end{align*}due to her weight. You may assume the mass of the web is negligible.
 If a small burst of wind sets her in motion, with what frequency will she oscillate?
 How many times will she go up and down in one s? In \begin{align*}20\;\mathrm{s}\end{align*}?
 How long is each cycle?
 Draw the \begin{align*}x\end{align*} vs \begin{align*}t\end{align*} graph of three cycles, assuming the spider is at its highest point in the cycle at \begin{align*}t=0\;\mathrm{s}\end{align*}.
Answers to Selected Problems
 a. \begin{align*}9.8 \times 10^5 \;\mathrm{N/m}\end{align*} b. \begin{align*}0.5 \;\mathrm{mm}\end{align*} c. \begin{align*}22 \;\mathrm{Hz}\end{align*}
 \begin{align*}3.2 \times 10^3 \;\mathrm{N/m}\end{align*}
 a. \begin{align*}16 \;\mathrm{Hz}\end{align*} b. \begin{align*}16\end{align*} complete cycles but \begin{align*}32\end{align*} times up and down, \begin{align*}315\end{align*} complete cycles but \begin{align*}630\end{align*} times up and down c. \begin{align*}0.063 \;\mathrm{s}\end{align*}
Investigation
 Your task: Match the period of the circular motion system with that of the spring system. You are only allowed to change the velocity involved in the circular motion system. Consider the effective distance between the block and the pivot to be to be fixed at 1m. The spring constant(13.5N/m) is also fixed. You should view the charts to check whether you have succeeded. Instructions: To alter the velocity, simply click on the Select Tool, and select the pivot . The Position tab below will allow you to numerically adjust the rotational speed using the Motor field. To view the graphs of their respective motion in order to determine if they are in sync, click on Chart tab below.

 Now the mass on the spring has been replaced by a mass that is twice the rotating mass. Also, the distance between the rotating mass and the pivot has been changed to 1.5 m. What velocity will keep the period the same now?
