AP Physics 1 Unit 7: Oscillations | Notes and practice

Notes

SHM happens when the restoring force is proportional to displacement and points back toward equilibrium: $F = -kx$.

Amplitude $A$: the maximum displacement. Period $T$: time for one cycle. Frequency $\displaystyle f = \frac{1}{T}$.
Position: $x = A\cos(2\pi ft)$.
At equilibrium: speed is maximum, acceleration and net force are zero.
At the extremes: speed is zero, acceleration and force are maximum.

Mass on a spring: $\displaystyle T_s = 2\pi\sqrt{\frac{m}{k}}$. It doesn't depend on amplitude or on $g$.
Simple pendulum (small angles): $\displaystyle T_p = 2\pi\sqrt{\frac{\ell}{g}}$. It doesn't depend on mass or amplitude.

Total mechanical energy $\displaystyle E = \frac{1}{2}kA^2$, constant without friction.
Energy moves back and forth: all potential at the extremes, all kinetic at equilibrium. $\displaystyle v_{max} = A\sqrt{\frac{k}{m}}$.
Doubling the amplitude makes the total energy four times as large.

Energy shifts between kinetic and potential, but the total stays the same.

A mass on a spring completes 10 cycles in 5.0 s. What is its frequency?

At which position does a mass on a spring have its greatest speed?

Quadrupling the mass on a spring makes the period:

What happens to a pendulum's period if the bob's mass is doubled?

A pendulum is taken to the Moon, where $g$ is smaller. Its period:

A spring with $k = 100$ N/m oscillates with amplitude 0.20 m. What is its total mechanical energy?

At maximum displacement, a mass in SHM has:

What length pendulum has a period of about 2.0 s on Earth? ($g = 9.8$ m/s²)

Write your answer first, then open the scoring guide.

Free-response question

A 0.50 kg block on a spring ($k = 200$ N/m) is pulled 0.10 m from equilibrium and released on a frictionless surface.

(a) Find the period of oscillation.

Show the answer and scoring guide

$\displaystyle T = 2\pi\sqrt{\frac{0.50}{200}} = 2\pi(0.05) \approx 0.31$ s.

Check off the points you earned:

(b) Find the block's maximum speed.

Show the answer and scoring guide

$\displaystyle \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2 \Rightarrow v_{max} = 0.10\sqrt{\frac{200}{0.50}} = 0.10(20) = 2.0$ m/s.

Check off the points you earned:

(c) The block is instead pulled 0.20 m. How do the period and maximum speed change? Explain.

Show the answer and scoring guide

The period stays the same, because it depends only on $m$ and $k$. The maximum speed doubles to 4.0 m/s, because $v_{max} \propto A$.

Check off the points you earned: