7.2 - Frequency and Period of SHM

Angular Frequency and Period

The period of simple harmonic motion is directly tied to the angular frequency

\omega

\omega = \frac{2\pi}{T}

where

\omega

is the angular frequency (rad/s) and

T

is the period (s). Since frequency and period are inverses:

f = \frac{1}{T}

you can also write angular frequency in terms of frequency:

\omega = 2\pi f

Period Equations for Oscillators

Mass-spring oscillator. For an ideal mass-spring system:

T = 2\pi\sqrt{\frac{m}{k}}

where

m

is the mass (kg) and

k

is the spring constant (N/m). This comes from the angular frequency relationship

\omega = \sqrt{\frac{k}{m}}

: substitute that into

\omega = \frac{2\pi}{T}

and you get the period formula.

Simple pendulum. For a simple pendulum:

T = 2\pi\sqrt{\frac{L}{g}}

where

L

is the pendulum length (m) and

g

is gravitational acceleration (m/s²).

Why the pendulum formula has no mass. In 7.1 the small-angle restoring force came out to

F \approx -\frac{mg}{L}s

. That’s the same form as

F = -kx

, with

\frac{mg}{L}

playing the role of

k

. Put that into the spring formula:

T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{m}{mg/L}} = 2\pi\sqrt{\frac{L}{g}}

The mass cancels. Doubling the bob’s mass doubles the restoring force, but it also doubles the inertia, so the acceleration at each point of the swing (and the period) doesn’t change.

On the AP equation sheet these show up as

T = \frac{2\pi}{\omega} = \frac{1}{f}

T_s = 2\pi\sqrt{\frac{m}{k}}

(s for spring), and

T_p = 2\pi\sqrt{\frac{\ell}{g}}

(p for pendulum, with

\ell

for the length). The new examples in this unit use

g = 10

m/s², the value the AP exam uses for numerical answers (see 1.3). Using 9.8 m/s² (the equation sheet value) or 9.81 m/s² isn’t penalized.

Small angle approximation. The pendulum equation only holds for small angles (about 15° or less). At bigger angles the motion stops being simple harmonic, because the restoring force

mg\sin\theta

is no longer proportional to

\theta

\sin\theta

grows more slowly than

\theta

, so the real restoring force is a bit weaker than the SHM model predicts and the real period is a bit longer. The error starts out small (the true period is about 0.4% longer at 15° and about 1.7% longer at 30°). For angles under 15°:

\sin(\theta) \approx \theta

and that approximation is what makes the pendulum period derivation work in the first place.

Alternative Frequency Equation

Since

f = \frac{1}{T}

, you can flip the pendulum period into a frequency equation:

f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

showing that frequency is inversely related to period.

Example Problem

A 1.5 m pendulum is pulled back and released to swing freely. Find (a) the period and frequency on Earth, (b) the period and frequency on Mars where

g = 3.71

m/s², and (c) how the frequencies compare.

Solution

(a) On Earth, with

g_{\text{Earth}} = 9.81

m/s²:

T_{\text{Earth}} = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.5}{9.81}} = 2.46 \text{ s}

f_{\text{Earth}} = \frac{1}{T} = \frac{1}{2.46} = 0.407 \text{ Hz}

so 0.407 cycles go by each second.

(b) On Mars:

T_{\text{Mars}} = 2\pi\sqrt{\frac{1.5}{3.71}} = 4.00 \text{ s}

f_{\text{Mars}} = \frac{1}{4.00} = 0.250 \text{ Hz}

(c) Comparing them. Notice that weaker gravitational acceleration gives a longer period, so period and

g

are inversely related:

\frac{T_{\text{Mars}}}{T_{\text{Earth}}} = \frac{4.00}{2.46} = 1.63

The period on Mars is 1.63 times longer. And since a longer period means a smaller frequency:

\frac{f_{\text{Mars}}}{f_{\text{Earth}}} = \frac{0.250}{0.407} = 0.614

The frequency on Mars is about 61% of Earth’s, so the pendulum completes fewer cycles per second there.

Example Problem: Mass-Spring Period

A 0.25 kg glider on a level air track is attached to a spring with

k = 16

N/m. Find (a) the angular frequency, period, and frequency, (b) the period if the glider is swapped for a 1.0 kg glider, and (c) the period if the original glider is attached to a spring with

k = 64

N/m instead.

Solution

(a) Find the angular frequency first, then use

T = \frac{2\pi}{\omega} = \frac{1}{f}

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{16 \text{ N/m}}{0.25 \text{ kg}}} = 8.0 \text{ rad/s}

T = \frac{2\pi}{\omega} = \frac{2\pi}{8.0 \text{ rad/s}} = 0.785 \text{ s} \approx 0.79 \text{ s} \qquad f = \frac{1}{T} = 1.3 \text{ Hz}

(b)

T_s \propto \sqrt{m}

. The mass went up by a factor of 4, so the period goes up by a factor of

\sqrt{4} = 2

T = 2(0.785 \text{ s}) = 1.6 \text{ s}

(c)

T_s \propto \frac{1}{\sqrt{k}}

. The spring is 4 times stiffer, so the period is cut in half:

T = \frac{0.785 \text{ s}}{2} = 0.39 \text{ s}

Notice what’s not in

T_s = 2\pi\sqrt{\frac{m}{k}}

: the amplitude and

g

. Pull the glider out farther and the period is still 0.79 s. Hang the same mass from the same spring vertically and the period is still 0.79 s, because gravity only shifts the equilibrium point (see 7.1).

Example Problem: Timing a Pendulum

A student counts 20 complete cycles (out and back) of a pendulum in 34.0 s. Find (a) the period, (b) the frequency and angular frequency, and (c) the length of the pendulum.

Solution

(a) The period is the time for one cycle:

T = \frac{34.0 \text{ s}}{20} = 1.70 \text{ s}

(b)

f = \frac{1}{T} = \frac{1}{1.70 \text{ s}} = 0.588 \text{ Hz} \qquad \omega = \frac{2\pi}{T} = \frac{2\pi}{1.70 \text{ s}} = 3.70 \text{ rad/s}

(c) Solve

T_p = 2\pi\sqrt{\frac{\ell}{g}}

for

\ell

\ell = \frac{gT^2}{4\pi^2} = \frac{(10 \text{ m/s}^2)(1.70 \text{ s})^2}{4\pi^2} = 0.73 \text{ m}

Timing many cycles and dividing beats timing just one: your reaction-time error (a few tenths of a second) gets spread over all 20 cycles.

Common Mistakes

Confusing frequency with angular frequency. $\omega = 2\pi f$ . They’re related but not the same quantity.

Using the pendulum equation at large angles. $T = 2\pi\sqrt{\frac{L}{g}}$ only works for small angles (about 15° or less).

Forgetting the inverse relationship. Period and frequency move opposite ways: when one goes up, the other goes down.

Mixing up the mass-spring and pendulum formulas. Mass-spring uses $m$ and $k$ ; the pendulum uses $L$ and $g$ . Note that the pendulum’s period doesn’t depend on mass at all.

Botching the substitution when deriving frequency. Flip the whole expression: $f = \frac{1}{2\pi\sqrt{\frac{L}{g}}} = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$ .

Thinking gravity changes a spring’s period. $T_s = 2\pi\sqrt{\frac{m}{k}}$ has no $g$ in it. The same mass on the same spring has the same period lying on a table, hanging vertically, or on the Moon.

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