Standing Waves I the same type o waves move through a common region and their requencies,, are the same then so are their wavelengths,. This ollows rom: v=. Since the waves move through a common region, the material they move through is the same, so their speeds must be same. With v and the same or the two waves, must also be the same or the two waves. We saw last class that in the region where such waves overlap their superposition creates a new wave having the same speed, requency and wavelength, but with an amplitude that ranges rom (A 1 +A 2 ) or ully constructive intererence to A 1 A 2 or ully destructive intererence, with A i the respective amplitude o each incident wave (and a phase oset o rom 0 to 2π).
I the two waves are counter propagating (i.e. move in opposite directions) a new phenomena can occur. They can generate what is called a standing wave. Simulation: http://www2.biglobe.ne.jp/~norimari/science/javaed/e-wave4.html To be deinitive we consider transverse waves propagating in opposite directions along a long string (though this happens in all types o waves). What characterizes the stationary wave is that it has positions along the string where the amplitude is always zero (called nodes) and positions hal way between the nodes where the string oscillates with large amplitude (called anti-nodes). The anti-nodes oten move too ast to see and are just a blur.
One way or such counter propagating waves to occur is to have a taught string tied o to a wall at one end and oscillated at the other end to launch a sinusoidal wave down it, y 1(,t) = Asin( ω t k) showing just orward wave At the tied o end the wave will relect and invert to generate the counter propagating wave (moving in the opposite direction). y 2(,t) = Asin( ω+ t k) showing just the relected wave
I we consider just these 2 propagating waves: y 1(,t) = Asin( ω t k) y 2(,t) = Asin( ω+ t k) Then their superposition gives y(,t) = y 1(,t) + y 2(,t) y(, t) = [ 2A sin(k) ] cos( ω t) (see GRR p.409) Which is not a traveling wave. The cosine part is a harmonic oscillation in time, but this is spatially modulated by the term in the square brackets which is an amplitude that changes with spatial position (since it depends on k). Showing just the original right going wave & the envelope o the resulting standing wave
y(, t) = [ 2A sin(k) ] cos( ωt) The cosine term just oscillates back and orth between ±1 as time progresses. = 0 The dashed green line shows the behavior o the sine. 4 2π 2π y() = sin(k) = sin( ) (recall k = ) Consider speciic points: At = : 4 2π π y( ) = sin( ) = sin( ) = + 1 4 4 2 So at = : y(,t) = 2Acos( ωt) oscillates between ±2A 4 4 giving an anti-node.
y(, t) = [ 2A sin(k) ] cos( ωt) 3 = 0,,,, 2π 4 2 4 y() = sin(k) = sin( ) 2π At = : y( ) = sin( ) = sin( π ) = 0 2 4 2 So at = : y(,t) = 2A(0)cos( ω t) = 0 Giving a node. 2 2 3 3 2π3 3π At = : y( ) = sin( ) = sin( ) = 1 4 4 4 2 3 So at = : 4 3 y(,t) = 2A( 1)cos( ωt) oscillates between 2A 4 giving an anti-node. ±
y(, t) = [ 2A sin(k) ] cos( ωt) 3 = 0,,,, 2π 4 2 4 y() = sin(k) = sin( ) 2π At = : y( ) = sin( ) = sin(2 π ) = 0 So at = : y(,t) = 2A(0)cos( ω t) = 0 Giving a node. So every hal wavelength is a node with anti nodes in between them, also separated by a hal wavelength.
y(, t) = [ 2A sin(k) ] cos( ωt) or two counter propagating waves each o amplitude A the amplitude o the standing wave is 2A. But i the system has low damping (low riction) then the relected wave will relect back o o the oscillated end again and then relect o the ied end again and so on, until there are many more than two waves adding to give a net amplitude or the standing wave that is much larger than the input wave oscillation. This is called resonance and is much like pumping your legs at just the right time when you are on a swing. By doing that at the right time (phase o the swing) the amplitude o the swing can get very large even though the energy you are pumping in is relatively small.
Now such standing wave resonance can t happen or just any wavelength (or requency). It can only happen i an integer number o hal wavelengths it between the two ends o the string. Because the input oscillation amplitude is so much smaller than the resonant amplitude at the antinodes we can to good approimation treat the system as i the oscillated (driven) end is also a node. Mathematically, this condition or resonance on a string o length L is, n = L, n = 1, 2, 3, 2 Solving or gives the wavelengths o the waves that can generate standing waves on a string o this length, =, n = 1, 2, 3, n
Now since, v=, v = So since the wavelengths o waves that can generate standing waves on a string o length L are, =, n = 1, 2, 3, n Then the requencies that can generate standing waves are, n = v v nv = = n, n = 1, 2, 3, Where recall that or waves on a string: String tension = v = µ linear density µ= mass length
The irst ew harmonic requencies that can generate standing waves: nv n =, n = 1, 2, 3, requency Designation v 1 = undamental or 1 st harmonic 2 2v = 2 nd harmonic 3 3v = 3 rd harmonic 4 4v = 4 th harmonic N = node A = anti-node
Eample The E string o a guitar has a length o 65 cm and is stretched to a tension o 82 N. It vibrates at a undamental requency o 329.63 Hz. What is the mass o the string? With undamental, n nv = with n = 1 or the undamental, so, So, 1 1 v = But v = µ µ = Squaring both sides 2 = µ 4L 1 2
= µ 4L 2 1 2 Solving or µ So µ= m = L 4L 2 2 1 4L 2 2 1 But µ= m L 82 N m = = 4L 4(0.65 m)(329.63 Hz) 2 2 1 4 m = 2.90 10 kg = 0.29 gm