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Vibrations on a String

Module by: Paul Padley. E-mail the author

Summary: Vibrations on a string give rise to waves and normal modes

Vibrations on a String

Figure 1
Figure 1 (String-Fragment-small.gif)

Consider the forces on a short fragment of string F y = T sin ( θ + Δ θ ) T sin θ F y = T sin ( θ + Δ θ ) T sin θ F x = T cos ( θ + Δ θ ) T cos θ F x = T cos ( θ + Δ θ ) T cos θ Assume that the displacement in y is small and T T is a constant along the stringthus θ θ and θ + Δ θ θ + Δ θ are smallthen F x 0 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 + ] F x T [ 1 ( θ + Δ θ ) 2 2 1 + θ 2 2 + ] or F x T θ Δ θ F x T θ Δ θ which is very small.On the other hand F y T [ θ + Δ θ θ + ] F y T [ θ + Δ θ θ + ] or F y T Δ θ F y T Δ θ which is not nearly as small. So we will consider the y 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 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 F y = m a y m = μ Δ x m = μ Δ x where μ μ is the mass density a y = 2 y t 2 a y = 2 y t 2 now have T 2 y x 2 Δ x = μ Δ x 2 y t 2 T 2 y x 2 Δ x = μ Δ x 2 y t 2 2 y x 2 = μ T 2 y t 2 2 y x 2 = μ T 2 y t 2 Note dimensions, get a velocity T μ = v 2 T μ = v 2 2 y x 2 = 1 v 2 2 y t 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 a string.

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

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

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

recall ν = ω / 2 π ν = ω / 2 π so ν n = n v 2 L = n 2 L ( T μ ) 1 2 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 we must have an integral number of half sine waves λ n = 2 L n λ n = 2 L n end up with f n ( x ) = A n sin ( 2 π x λ n ) 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 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 n = n π L ( T μ ) 1 2 = n π L v = n ω 1 ω 1 ω 1 is the fundamental frequency

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