Cause : resonance.

**Standing transverse waves**

Form: sine .

Occurs in stringed instruments

Amplitude varies along the string. They all the vibrate with the same frequency.

Nodes (no vibrations) and Antinodes (amplitude maximal value)

Nodes at the fixed ends and between the antinodes.

Between two nodes vibrate the points with equal phase

Δφ = 0 ( points between two nodes) or 0.5 (points on either side of a node)

There is also:

v = f λ

- v speed of the travelling wave in the string
- λ length of one sine

**Standing waves in a string fixed at two ends **

**Example**

A string with length of 60 cm is fixed at both sides

The fundamental has a frequency of 400 Hz.

Calculate the speed of the wave in the string.

The distance from node to node N-N = ½ λ So 0.5 λ_{1}= 0.60 m λ_{1}_{ }= 1.20 m

λ_{f}_{ }is the wavelength of the fundamental mode

v = f_{1}_{ } λ_{1 }= (400)( 1.20) = 480 m/s

**Example**

Find now the frequency of the second harmonic

v = f_{2 }λ_{2}

λ_{2 }= 0.60 m

480 = f_{2 }0.60 f_{2}= 800 Hz

The frequency of the third harmonic is 1200 Hz ( 1.5 λ_{3} = 0.60 m

λ_{3}= _{ }0.40 m)

**Standing longitudinal waves**

Wind instruments and organ pipes

The same type of calculations

An antinode is created at the mouth of the pipe!

**Example**

Find the frequency of the fundamental of an** open** organ pipe with a length of 2.0 m

at a temperature of 20 ⁰C (speed of the sound is 343 m/s)

**Open** organ pipe : the open top is also an antinode

Distance from antinode to antinode A-A = ½ λ_{f}

So λ_{f }= (2)( 2.0) = 4.0 m

v = 343 m/s

v = f_{f} λ_{f}

343 = f_{f . }4.0

f_{f }= 85.8 Hz

**Example**

Find now the frequency of the fundamental of a** stopped **organ pipe with the same length

Fundamental:

Stoppedorgan pipe : the closed top created a node

length pipe (N-A) = ¼ λ So λ= (4)( 2.0) = 8.0 m

v = f_{f }_{ }λ_{f} 343 = f_{f} 8.0 f_{f}= 42.9 Hz ( 2 x lower then the f_{f }of an open pipe)

**Example**

Calculate the frequency of the second harmonic of a stopped organ pipe with the same length of 2.0 m at the same temperature of 20 ⁰C

length organ pipe = 2.0 m

length = ¾ λ _{1} λ_{1}= 2.67 m

f_{1} = 128.5 Hz