Learning�Object

Standing�Waves:�Resonant�Frequencies

A�standing�wave�oscillates�on�a�string�of�length�L.�The�ends�are��ixed�and�the�wavelength�is�1/3�L.�Which�envelope�below�most�accurately�represents�this�standing�wave?�

A.

B.

C.

D.

E.

Learning�Object

Standing�Waves:�Resonant�Frequencies

Solution�

C.

ReferenceR.�Hawkes�et�al.,�Physics�for�Scientists�and�Engineers:�An�Interactive�Approach,�Revised�Custom�Vol�1.�(Nelson,�Toronto,�CA,�2015),�pp.�409-413.

Method�1�-�Number�of�Wavelengths�in�L

The�string�is�length�L�and�the��standing�wave's�wavelength�is�1/3�L.�This�means�that�the�wave�must�have�3�wavelenths�from�one��ixed�end�to�the�other.�Lets�examine�the�options�to�see�which�one��its�this�condition.�Each�wave�is�shown�as�its�envelope—the�outline�of�maximum�displacement�of�each�particle,�respectively.�Wave�A�has�2.5�wavelenths�throughout�the�length�of�the�string.�This�is�close,�but�it�is�not�what�we�are�after.�Wave�B�has�2�wavelenths�in�L—also�does�not�correspond.�Wave�C�has�3�wavelenths�in�L.�Does�this�satisfy�1/3�L?�Yes—wave�C�has�a�wavelength�of�1/3�L.�For�completion,�the�wavelengths�of�wave�D�and�wave�E�are�shorter�and�longer�than�1/3�L,�respectively,�so�they�do�not�match.

Method�2�-�Antinodes

We�can�also�solve�this�problem�by�using�the�following�equation�to�determine�the�number�of�antinodes�in�our�wave�of�interest:

wavelength�=�2�L�/�m�=>�m�=��2�L�/�wavelength��

Where�m�is�a�non-zero�positive�integer�and�equal�to�the�number�of�antinodes,�and�L�is�the�length�of�the�string.�Before�we�tackle�this�approach.�Let’s�review�nodes�and�antinodes.�Particles�in�a�standing�wave�undergo�simple�harmonic�motion.�Each�particle’s�amplitude�is�in�the�range�of�zero�to�some�maximum�magnitude.�Simply�put,�the�points�on�a�standing�wave�on�a�string�with��ixed�ends�that�have�zero�amplitude�are�called�nodes,�and�the�points�that�have�maximum�amplitude�are�called�antinodes�(a�helpful�mnemonic:�‘a’�for�amplitude�and�antinode).�Subbing�the�wavelength,�1/3�L,�into�the�above�equation�we�get:�

m�=�(2�/�L)�/�(1/3�L)�

Cancelling�L�and�solving�for�m�gives:

m�=�6

We�learn�that�our�wave�of�interest�has�6�antinodes.�After�examining�the�envelopes,�we�determine�that�wave�A,�B,�C,�D�and�E�have�5,�4,�6,�7�and�3�points�of�maximum�displacement/amplitude,�respectively�(remember�that�the�envelope�shows�the�outline�of�maximum�displacement�of�each�particle—we�count�the�positive/negative�maxima�as�a�single�antinode).�As�we�just�learned,�these�points�are�antinodes.�Therefore,�wave�C�is�correct.�

Antinode:�ANode:�N

A

N

Wavelength:�1/3�L

Lenth:�L