Standing Waves in Air Columns - Learn

Standing Waves in Air Columns

Standing waves also form in air columns. The compressions and rarefactions of the sound wave are reflected by both open and closed ends of a tube. This can allow resonance to occur in the tube and the length of the tube determines the frequency of the resonating sounds.

Two situations are considered:

An air column with openings at both ends
An air column with an opening at one end and the other end closed

Nodes form at positions where there is a closed end and antinodes form at positions where there is an opening.

Many of the same definitions from standing waves in strings apply here:

Standing or stationary waves in air columns vibrate at the resonant frequencies of the air column. The resonant frequencies produced by these vibrations which produce standing waves are known as harmonics. The simplest form of vibration is called the fundamental frequency or first harmonic. The other modes of vibration are known as the second harmonic, third harmonic and so on. Other than the first harmonic, all other harmonics are known as overtones.

The resonant frequencies or harmonics in an air column can be determined by the relationship between the length of the air column $l$ , and the wavelength $\lambda$ , of the corresponding standing wave.

For open air columns:

$\lambda =\cfrac { 2l }{ n }$

where:

$\lambda$ = the wavelength in metres (m)

$l$ = the length of the air column in metres (m)

$n$ = the number of harmonics, which is also the number of antinodes (1, 2, 3, 4 etc)

The relationship between $\lambda$ , $l$ and $n$ is illustrated below:

Another equation derives from the relationship: $v=f\lambda$ . This equation allows the frequency of the standing wave to be determined:

$f=\cfrac { nv }{ 2l }$

where:

$f$ = the frequency of the wave in Hertz (Hz)

$v$ = the velocity of the wave in ${ m }/{ s }$

$l$ = the length of the string in metres (m)

$n$ = the number of harmonics, which is also the number of antinodes (1, 2, 3, 4 etc)

For air columns closed at one end:

$\lambda =\cfrac { 4l }{ n }$

$f=\cfrac { nv }{ 4l }$

Note: only odd harmonics exist for closed air columns.

Example 1:

An open-ended pipe 54cm in length has sound travelling in it at $320{ m }/{ s }$ .

a) What is the wavelength of the fourth harmonic?

b) What is the frequency of the fourth harmonic?

Answers:

a) using: $\lambda =\cfrac { 2l }{ n }$

$\lambda =\cfrac { 2\times 0.54 }{ 4 }$

$\lambda =0.27m$

b) using: $f=\cfrac { nv }{ 2l }$

$f=\cfrac { 4\times 320 }{ 2\times 0.54 }$

$f=1185.2\;Hz$

Example 2:

An closed-ended pipe has sound travelling in it at $320{ m }/{ s }$ . It produces a fundamental frequency of 150Hz.

a) What is the length of the air column?

b) What is the length of the next possible harmonic above the fundamental frequency produced by this air column?

Answers:

a) using: $f=\cfrac { nv }{ 4l }$

$150=\cfrac { 1\times 320 }{ 4l }$

$l=\cfrac { 1\times 320 }{ 4\times 150 }$

$l=53.3\;cm$

b) using: $\lambda =\cfrac { 4l }{ n }$ and n=3 because only odd number harmonics can exist for a closed air column:

$\lambda =\cfrac { 4l }{ n }$

$\lambda =\cfrac { 4\times 53.3 }{ 3 }$

$\lambda =71.1\;cm$

Colliding Objects (1-D)

When objects collide, their momentum changes. Colliding objects are goverened by the law of conservation of momentum, stated as follows:

Law of Conservation of Momentum:

Provided no external forces act on the system, the total momentum of the system before any collision occurs is equal to the total momentum of the system after the collision

We can apply this law of collisions to determine the formula:

Total momentum before collision = Total momentum after collision

Or stated mathematically:

${ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }$

Where

${ m }_{ 1 }$ = mass of object 1 in kg

${ u }_{ 1 }$ = initial velocity object 1 in $\cfrac { m }{ s }$

${ v }_{ 1 }$ = final velocity object 1 in $\cfrac { m }{ s }$

${ m }_{ 2 }$ = mass of object 2 in kg

${ u }_{ 2 }$ = initial velocity object 2 in $\cfrac { m }{ s }$

${ v }_{ 2 }$ = final velocity object 2 in $\cfrac { m }{ s }$

Examples of collisions that we will consider:

A moving object collides with a stationary object and then stops
Objects moving in the same direction, colliding and sticking together
Objects moving in the same direction, colliding and moving in the same direction or rebounding
Objects moving in the opposite direction, colliding and moving in the same direction or rebounding

As momentum is a vector, in the equation above, initial and final velocities must take into account direction.

Example 1: A moving object collides with a stationary object and then stops

A car with a mass of $500kg$ travelling at $3\cfrac { m }{ s }$ to the north, collides with a $750kg$ car which is stationary. After the collision the $500kg$ car stops. What happens to the $750kg$ car?

${ m }_{ 1 }$ = $500kg$

${ u }_{ 1 }$ = $3\cfrac { m }{ s }$ to the north (note: north as positive)

${ v }_{ 1 }$ = $0\cfrac { m }{ s }$

${ m }_{ 2 }$ = $750kg$

${ u }_{ 2 }$ = $0\cfrac { m }{ s }$

${ v }_{ 2 }$ = ?

Substituting values into:

${ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }$

$(500\times 3)+0=0+(750\times { v }_{ 2 })$

$1500=750{ v }_{ 2 }$

${ v }_{ 2 }=2\cfrac { m }{ s }$ north

Example 2: Objects moving in the same direction, colliding and sticking together

A shopping cart with a mass of $40kg$ travelling at $1.5\cfrac { m }{ s }$ east, collides with a $10kg$ shopping cart travelling at $2.5\cfrac { m }{ s }$ in the same direction. After the collision the trolleys were stuck together and moving at the same speed. What is the velocity of the combined masses after the collision.

${ m }_{ 1 }$ = $40kg$

${ u }_{ 1 }$ = $1.5\cfrac { m }{ s }$ to the east (note: east as positive)

${ v }_{ 1 }$ = ?

${ m }_{ 2 }$ = $10kg$

${ u }_{ 2 }$ = $2.5\cfrac { m }{ s }$

${ v }_{ 2 }$ = ?

In problems where the objects join together after the collision, it is convenient to use the combined mass. We will write this as ${ m }_{ 3 }$ :

${ m }_{ 3 }$ = $40+10=50kg$

${ v }_{ 3 }$ = ?

Substituting values into:

${ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 3 }{ v }_{ 3 }$

$(40\times 1.5)+(10\times 2.5)=(50\times { v }_{ 3 })$

$85=50{ v }_{ 3 }$

${ v }_{ 3 }=1.7\cfrac { m }{ s }$ east

Example 3: Objects moving in the same direction, colliding and moving in the same direction or rebounding

A $250g$ ice puck travelling at $5\cfrac { m }{ s }$ east, collides with an identical ice puck travelling in the same direction at $2.5\cfrac { m }{ s }$ . After the collision the first ice puck rebounds in the opposite direction at $1.5\cfrac { m }{ s }$ . What is the velocity of the second ice puck after the collision?

${ m }_{ 1 }$ = $250g$

${ u }_{ 1 }$ = $5\cfrac { m }{ s }$ to the east (note: east as positive)

${ v }_{ 1 }$ = $1.5\cfrac { m }{ s }$ west = $-1.5\cfrac { m }{ s }$

${ m }_{ 2 }$ = $250g$

${ u }_{ 2 }$ = $2.5\cfrac { m }{ s }$ to the east

${ v }_{ 2 }$ = ?

Substituting values into:

${ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }$

$(0.25\times 5)+(0.25\times 2.5)=(0.25\times -1.5)+(0.25\times { v }_{ 2 })$

$1.25+0.625=-0.375+0.25{ v }_{ 2 }$

$1.875=-0.375+0.25{ v }_{ 2 }$

$2.25=0.25{ v }_{ 2 }$

${ v }_{ 2 }=9\cfrac { m }{ s }$ west

Example 4: Objects moving in the opposite direction, colliding and moving in the same direction or rebounding

A ball with a mass of $2kg$ travelling at $3\cfrac { m }{ s }$ to the north, collides with another ball with a mass of $4kg$ moving in the opposite direction at $3\cfrac { m }{ s }$ . After the collision the $2kg$ ball bounces in the opposite direction at $4\cfrac { m }{ s }$ . What happens to the $4kg$ ball?