The oscillating object does not lose any energy in SHM. Friction is assumed to be zero.

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*}

The frequency, \begin{align*}f,\end{align*}*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.

\begin{align*} T = \frac{1}{f}\end{align*}

\begin{align*} T_{\text{spring}} = 2\pi\sqrt{\frac{m}{k}}\end{align*}

\begin{align*} F_{sp} = -kx \end{align*}

\begin{align*} U_{sp} = \frac{1}{2} kx^2 \end{align*}

### Simulation

Mass & Springs (PhET Simulation)

### Review

- A rope can be considered as a spring with a very high spring constant \begin{align*}k,\end{align*}
k, 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*}
k of a rope that stretches by \begin{align*}1\;\mathrm{mm}\end{align*}1mm when a \begin{align*}100\;\mathrm{kg}\end{align*}100kg weight hangs from it? - If a boy of \begin{align*}50\;\mathrm{kg}\end{align*}
50kg 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?

- What is the \begin{align*}k\end{align*}
- If a \begin{align*}5.0\;\mathrm{kg}\end{align*}
5.0kg mass attached to a spring oscillates 4.0 times every second, what is the spring constant \begin{align*}k\end{align*}k of the spring? - A horizontal spring attached to the wall is attached to a block of wood on the other end. All this is sitting on a frictionless surface. The spring is compressed \begin{align*}0.3\;\mathrm{m}\end{align*}
0.3m . Due to the compression there is \begin{align*}5.0\;\mathrm{J}\end{align*}5.0J of energy stored in the spring. The spring is then released. The block of wood experiences a maximum speed of \begin{align*}25\;\mathrm{m/s}\end{align*}25m/s .- Find the value of the spring constant.
- Find the mass of the block of wood.
- What is the equation that describes the position of the mass?
- What is the equation that describes the speed of the mass?
- Draw three complete cycles of the block’s oscillatory motion on an \begin{align*}x\end{align*}
x vs. \begin{align*}t\end{align*}t graph.

- A spider of \begin{align*}0.5\;\mathrm{g}\end{align*}
0.5g walks to the middle of her web. The web sinks by \begin{align*}1.0\;\mathrm{mm}\end{align*}1.0mm 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*}
20s ? - How long is each cycle?
- Draw the \begin{align*}x\end{align*}
x vs \begin{align*}t\end{align*}t graph of three cycles, assuming the spider is at its highest point in the cycle at \begin{align*}t=0\;\mathrm{s}\end{align*}t=0s .

### Review (Answers)

- a. \begin{align*}9.8 \times 10^5 \;\mathrm{N/m}\end{align*}
9.8×105N/m b. \begin{align*}0.5 \;\mathrm{mm}\end{align*}0.5mm c. \begin{align*}22 \;\mathrm{Hz}\end{align*}22Hz - \begin{align*}3.2 \times 10^3 \;\mathrm{N/m}\end{align*}
3.2×103N/m - a. \begin{align*}110 \;\mathrm{N/m}\end{align*}
110N/m d. \begin{align*}v(t)=(25) \cos(83\mathrm{t})\end{align*}v(t)=(25)cos(83t) - a. \begin{align*}16 \;\mathrm{Hz}\end{align*}
16Hz b. \begin{align*}16\end{align*}16 complete cycles but \begin{align*}32\end{align*}32 times up and down, \begin{align*}315\end{align*}315 complete cycles but \begin{align*}630\end{align*}630 times up and down c. \begin{align*}0.063 \;\mathrm{s}\end{align*}0.063s

### Explore More

IMatch 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?