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Vibrating String

VIBRATING STRINGS
Vibrating strings can be modeled by the wave equation. We can treat a string as a one dimensional body in x, with a displacement u (a function of both space and time).
(1) (∂2u(x, t))/(∂t2) = C2(∂2u(x, t))/(∂x2)
If this equation models the motion of a pulse in a string then u is displacement,t is time, x is axial coordinate, (Fig. 1↓). Both sides of the equality should have the same units, therefore C should have velocity units. Recall that in the wave equation, C is equivalent to the “speed of sound” and so will describe the speed at which a disturbance propagates along the string.
figure images/string1.png
Figure 1 An example of the displacement of a string at a given time.
DERIVATION OF WAVE EQUATION ON A VIBRATING STRING
Figure 2↓ shows a very small piece of a string under tension. In this derivation, we will make following assumptions:
figure images/string.png
Figure 2 A very small piece of a string under tension
  • Displacements are small.
  • The tension (T) in the spring is constant, this is the same as assuming the dominant motion is in y direction.
In the following equations, u is the displacement in y direction. m is the mass per unit of length, T is the constant tension in the string. Sine the dominant motion is in y direction, one needs to concentrate on equilibrium equations in this direction. Newton’s second law should be applied to this small piece of string in y direction.
(2) ⎲⎳Fy = mdx(∂2u(x, t))/(∂t2)
(3)  − Tsin(θ) + Tsin(θ + (∂θ)/(∂x)dx) = mdx(∂2u(x, t))/(∂t2)
Since θ is small, by using taylor series and neglecting higher order terms, one can approximate sin(θ) by θ and sin(θ + (∂θ)/(∂x)dx) by θ + (∂θ)/(∂x)dx so
(4)  − Tθ + Tθ + T(∂θ)/(∂x)dx = mdx(∂2u(x, t))/(∂t2)
on the other hand from the geometry of the problem θ = (∂u)/(∂x). Finally one has
(5) T(∂2u(x, t))/(∂x2) = m(∂2u(x, t))/(∂t2)
(6) (∂2u(x, t))/(∂t2) = (T)/(m)(∂2u(x, t))/(∂x2)
Comparing 6↑ with 1↑, we see that the speed at which a disturbance progagates along the string, C, is:
(7) C = √((T)/(m))