Standing Wave on a StringJessica Weng

PHYS 101 LO6

How is standing wave created?When you pluck a string with both ends fixed: waves travel back and forth along it

waves get reflected by the fixed end creating wave travelling in opposite direction

Results in the superposition of two waves both having the same wavelength, frequency, and amplitude but travelling in opposite direction

Refresher: Nodes and AntinodesNode:

where amplitude = 0 Antinode:

where amplitude is maximum

Review QuestionHow many nodes are present in the following standing waves?

Review Question (cont’d)Solution:

1) 0 node2) 1 node3) 3 nodes4) 6 nodes

Standing Wave & StringIf we have a string of length L,

starting at x= 0 and ends at x=L, with both ends fixed: Only certain wavelengths will be able to fit on the string in order to produce standing waveWhich wavelengths???

Let’s do some math to find out.

Standing Wave & String (cont’d)The following equation describes the

amplitude of standing wave:

For our string fixed at both ends x= 0 and x =L:Amplitude = 0 at the two ends

The argument of sin must 0

Standing Wave & String (cont’d)Sin(2pi/λ*L) = 0 2pi/λ*L = m*pi m = a positive non-

zero integerRearrange the equation for

wavelength:λ=2L/m, m=1,2,3,4… λ=2L, L,2L/3,L/2….

These are the wavelengths that a string with both ends fixed can oscillate with in a standing wave patternnormal modes of vibration

Standing Wave & String (cont’d)Frequencies corresponding to normal

modes of vibration: v=λf rearrangef=v/λ =v/(2L/m) = m/2L *v

substitute v = (T / μ)^1/2

Standing Wave & String (cont’d)Fundamental frequency / first

harmonic:Lowest frequency & longest wavelength λ=2L

Higher frequencies have higher m values and are integer multiples of first harmonicfm=mf1

A conceptual questionHow many nodes are present

between the fixed ends of a string vibrating in at sixth harmonic?

A conceptual question (cont’d)Hint:

Find its frequencyFind its number of antinodesThink about the relationship between antinodes and nodes

A conceptual question (cont’d)Solution:

A string at sixth harmonic:fm = mf1 f6=6*f1m=6 there are six antinodes# of nodes = # of antinodes – 1 = m-1 =5

Another conceptual questionWhen the string player puts a finger

down tightly on the string:1.How has the part of the string that vibrates changed?2.How does this change the sound waves that the string makes?3.How does this change the sound that is heard?

Another conceptual question (cont’d)Hint:

Think about what happens to L in the equation!

Another conceptual question (cont’d)Solution:

1.The part of the string that can vibrate becomes shorter as the finger becomes the new fixed end of the string.2.The new sound wave is shorter, so its frequency is higher.3. It sounds higher / it has a higher pitch. (because of the higher frequency)

A practical questionWe have a stretched string of length

10m with both ends fixed. Its frequency at its fourth harmonic is 240Hz. 1)What is the longest wavelength standing wave possible on this string? 2)What is its fundamental frequency?

A practical question(cont’d)Hint:

We are given value of L: Wavelength = 2L/mAt longest wavelength, how many antinodes do we have?

We are given frequency at fourth harmonic:fm=m*f1How many antinodes do we have at fourth harmonic?

A practical question (cont’d)Solution:

1)Wavelength = 2L/m at longest wavelength, there is 1 antinode, so m=1L=10mWavelength = 2*(10m) / 1 = 20m = longest

wavelength2)fm=m*f1at fourth harmonic, there are four antinodes, so m

=4f4=240Hzf4=4*f1f1=f4/4=240Hz/4=60Hz = fundamental frequency

Thank you for watching!