Mechanical oscillations

Oscillatory motion, amplitude, period and frequency, free and forced oscillations, thread and spring pendulums, resonance.

Terms

Studied 0 of 9

Formulas

Period of oscillations

T = \dfrac{t}{N}

Quantities
Symbol	Name	Unit
$T$	period of oscillations	s
$t$	time of the oscillations	s
$N$	number of full oscillations in time $t$	—

Solved for another quantity:

t = T\,N

N = \dfrac{t}{T}

Note. Example: in $t = 20\ \text{s}$ a pendulum made $N = 40$ oscillations, so $T = \dfrac{20}{40} = 0.5\ \text{s}$ . To measure the period more accurately, many oscillations are counted.

Frequency of oscillations

\nu = \dfrac{1}{T}

Quantities
Symbol	Name	Unit
$\nu$	frequency of oscillations	Hz
$T$	period of oscillations	s

Solved for another quantity:

T = \dfrac{1}{\nu}

\nu = \dfrac{N}{t}

Note. Frequency and period are reciprocal quantities. Example: for $T = 0.25\ \text{s}$ the frequency is $\nu = \dfrac{1}{0.25} = 4\ \text{Hz}$ ; for $\nu = 50\ \text{Hz}$ the period is $T = \dfrac{1}{50} = 0.02\ \text{s}$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Thread pendulum

The load is moved aside by the amplitude and released. The red marks are the extreme positions (displacement ±A), the dashed line is the equilibrium position. The scene is not to scale.

t

= 0 sPlayback speed

Graphs

The graph of the displacement from the equilibrium position is shown for four periods. Click the graph to move through it.

x(t)

A

5 cm

T

1.99 s

\nu

0.5 Hz

N

x

5 cm

Formulas with your numbers

T = \dfrac{t}{N} = \dfrac{7.95}{4} = 1.99\ \text{s}

\nu = \dfrac{1}{T} = \dfrac{1}{1.99} = 0.5\ \text{Hz}

Additional (the period for small deflections; usually not needed in OGE calculation problems):

T = 2\pi\sqrt{\dfrac{l}{g}} = 2\pi\sqrt{\dfrac{1}{10}} = 1.99\ \text{s}

How to read the graphs

The $x(t)$ graph is a smooth wave-like curve: the body goes to one side ( $x > 0$ ), then to the other ( $x < 0$ ), passing the equilibrium position ( $x = 0$ ).
The amplitude $A$ is the greatest value of $x$ on the graph (the height of a crest above the time axis). A greater amplitude makes the curve "taller" while its length along the time axis stays the same.
The period $T$ is the distance along the time axis between two neighbouring crests (or two neighbouring troughs): in that time the body makes one full oscillation and returns to the same state. The greater the period, the slower the oscillations.
The number of crests on the graph in time $t$ is the number of oscillations $N$ ; then $T = t / N$ , and the frequency $\nu = 1 / T$ shows how many crests fall into $1\ \text{s}$ .
In one period the body covers the distance $4A$ (from one extreme to the other and back), but its displacement over the period is zero — it returns to the starting point.

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Terms

Studied 0 of 9

Formulas

Period of oscillations

T = \dfrac{t}{N}

Quantities
Symbol	Name	Unit
$T$	period of oscillations	s
$t$	time of the oscillations	s
$N$	number of full oscillations in time $t$	—

Solved for another quantity:

t = T\,N

N = \dfrac{t}{T}

Note. Example: in $t = 20\ \text{s}$ a pendulum made $N = 40$ oscillations, so $T = \dfrac{20}{40} = 0.5\ \text{s}$ . To measure the period more accurately, many oscillations are counted.

Frequency of oscillations

\nu = \dfrac{1}{T}

Quantities
Symbol	Name	Unit
$\nu$	frequency of oscillations	Hz
$T$	period of oscillations	s

Solved for another quantity:

T = \dfrac{1}{\nu}

\nu = \dfrac{N}{t}

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Thread pendulum

The load is moved aside by the amplitude and released. The red marks are the extreme positions (displacement ±A), the dashed line is the equilibrium position. The scene is not to scale.

t

= 0 sPlayback speed

Graphs

The graph of the displacement from the equilibrium position is shown for four periods. Click the graph to move through it.

x(t)

A

5 cm

T

1.99 s

\nu

0.5 Hz

N

x

5 cm

Formulas with your numbers

T = \dfrac{t}{N} = \dfrac{7.95}{4} = 1.99\ \text{s}

\nu = \dfrac{1}{T} = \dfrac{1}{1.99} = 0.5\ \text{Hz}

Additional (the period for small deflections; usually not needed in OGE calculation problems):

T = 2\pi\sqrt{\dfrac{l}{g}} = 2\pi\sqrt{\dfrac{1}{10}} = 1.99\ \text{s}

How to read the graphs

The

x(t)

graph is a smooth wave-like curve: the body goes to one side (

x > 0

), then to the other (

x < 0

), passing the equilibrium position (

x = 0

The amplitude

A

is the greatest value of

x

on the graph (the height of a crest above the time axis). A greater amplitude makes the curve "taller" while its length along the time axis stays the same.

The period

T

is the distance along the time axis between two neighbouring crests (or two neighbouring troughs): in that time the body makes one full oscillation and returns to the same state. The greater the period, the slower the oscillations.

The number of crests on the graph in time

t

is the number of oscillations

N

; then

T = t / N

, and the frequency

\nu = 1 / T

shows how many crests fall into

1\ \text{s}

In one period the body covers the distance

4A

(from one extreme to the other and back), but its displacement over the period is zero — it returns to the starting point.