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Vibrations on a string give rise to waves and normal modes

Vibrations on a string

Consider the forces on a short fragment of string F y = T sin ( θ + Δ θ ) T sin θ F x = T cos ( θ + Δ θ ) T cos θ Assume that the displacement in y is small and T is a constant along the stringthus θ and θ + Δ θ are smallthen F x 0 we can see this by expanding the trig functions F x T [ 1 ( θ + Δ θ ) 2 2 1 + θ 2 2 + ] or F x T θ Δ θ which is very small.On the other hand F y T [ θ + Δ θ θ + ] or F y T Δ θ which is not nearly as small. So we will consider the y component of motion, but approximate there is no x component F y = T sin ( θ + Δ θ ) T sin θ T tan ( θ + Δ θ ) T tan θ = T ( y ( x + Δ x ) x y x ) = T 2 y x 2 Δ x Also we can write: F y = m a y m = μ Δ x where μ is the mass density a y = 2 y t 2 now have T 2 y x 2 Δ x = μ Δ x 2 y t 2 2 y x 2 = μ T 2 y t 2 Note dimensions, get a velocity T μ = v 2 2 y x 2 = 1 v 2 2 y t 2 The second space derivative of a function is equal to the second time derivative of a function multiplied by a constant.

Normal modes on a string

Before considering traveling waves, we are going to look at a special case solution to the wave equation. This is the case of stationary vibrations of astring.

For example here, lets consider the case where both ends of the string are fixed at y = 0 . Now we vibrate the string. Every point along the string acts like a littledriven oscillator. So lets assume that every point on string has a time dependence of the form cos ω t and that the amplitude is a function of distance Assume y ( x , t ) = f ( x ) cos ω t then 2 y t 2 = ω 2 f ( x ) cos ω t 2 y x 2 = 2 f x 2 cos ω t Substitute into wave equation 2 y x 2 = 1 v 2 2 y t 2 2 f x 2 cos ω t = ω 2 v 2 f ( x ) cos ω t Then every f ( x ) that satisfies: 2 f x 2 = ω 2 v 2 f is a solution of the wave equation

A solution is (requiring f ( 0 ) = 0 since ends fixed) f ( x ) = A sin ( ω x v ) Another boundary condition is f ( L ) = 0 so get A sin ( ω L v ) = 0 Thus ω L v = n π ω = n π v L

Be careful with the equations above: v is the letter vee and is for velocity. now we introduce the frequency ν which is the Greek letter nu.

recall ν = ω / 2 π so ν n = n v 2 L = n 2 L ( T μ ) 1 2 This is a very important feature of wave phenomena. Things can be quantized. This is why a musical instrument will play specific notes. Note, that wemust have an integral number of half sine waves λ n = 2 L n end up with f n ( x ) = A n sin ( 2 π x λ n ) leading to y n ( x , t ) = A n sin ( 2 π x λ n ) cos ω n t where ω n = n π L ( T μ ) 1 2 = n π L v = n ω 1 ω 1 is the fundamental frequency

Questions & Answers

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Source:  OpenStax, Waves and optics. OpenStax CNX. Nov 17, 2005 Download for free at http://cnx.org/content/col10279/1.33
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