The wave equation
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The document discusses the wave equation and its application to modeling vibrating strings and wind instruments. It describes how the wave equation can be separated into independent equations for time and position using the assumption that displacement is the product of separate time and position functions. This separation leads to trigonometric solutions that satisfy the boundary conditions of strings fixed at both ends. The solutions represent standing waves with discrete frequencies determined by the length, tension, and density of the string. Similar methods apply to wind instruments with different boundary conditions.
The wave equation
- 1. 1 The Wave Equation SPECIAL TOPICS: PARTIAL DIFFERENTIAL EQUATIONS Dhaval Jalalpara A.
- 2. 2 Sub:- Maths Division:- A Topic:- Wave Equation
- 3. 3 Motion of a string Imagine that a stretched string is vibrating. The wave equation says that, at any position on the string, acceleration in the direction perpendicular to the string is proportional to the curvature of the string. x u displacement =u(x,t)
- 4. 4 The one-dimensional wave equation Let • x = position on the string • t = time • u(x, t) = displacement of the string at position x and time t. • T = tension (parameter) • ρ = mass per unit length (parameter) Then Equivalently, ρ ∂2 ∂t2 u(x,t)=T ∂2 ∂x2 u(x,t) utt=a2 uxx wherea=Tρ
- 5. 5 Solving the one-dimensional WE First, we make a wild assumption: suppose that u is a product of a function of x and a function of t: Then the wave equation becomes So, The only way for this equation to be true for all x and for all t is for both sides to be constant; that is, T’’(t)/a2 T(t) = λ and X”(x)/X(x) = λ. u(x,t)=X(x)T(t) X(x)′′T(t)=a2 ′′X(x)T(t) ′′T (t) a2 T(t) = ′′X(x) X(x)
- 6. 6 Separation of variables Now means that If λ < 0, T(t) and X(x) are both trig functions.* ′′T(t) a2 T(t) = ′′X(x) X(x) =λ ′′T(t)−λa2 T(t)=0,and ′′X(x)−λX(x)=0, *If λ = 0, they’re linear, and if λ > 0, they’re exponential.
- 7. 7 Clarification: the sign of λ There are three types of solutions to the equation 1. if λ = 0, then X(x) is linear (ax + b), which won't satisfy the boundary conditions; 2. if λ > 0, X(x) is exponential (ke√λt ), which also won't satisfy the boundary conditions; and 3. if λ < 0, then X(x) is a sinusoid that satisfies the boundary conditions. ′′X(x)−λX(x)=0 X(x)=ksin −λ( )x( )
- 8. 8 Boundary conditions In a stringed instrument, each end of the string is fixed; if the string has length L, then, for all t, Since T(t) ≠ 0 for all t, X(0) = X(L) = 0; thus, X(x) could be a sum of sine terms with zeros at x = L: u(0,t)=X(0)T(t)=0,and u(L,t)=X(L)T(t)=0. X(x)=ksinnπx/L( ) 0 L
- 9. 9 Finding T(t) Now we know two things: This means that Now substitute λ into the equation for T: And find the solution: X(x)=ksinnπx/L( )′′X(x)−λX(x)=0 λ=− n2 π2 L2 ′′T(t)−λa2 T(t)=0 T(t)=bsin anπt L +ccos anπt L
- 10. 10 Putting it all together We now have a family of solutions for the wave equation (one for every n): Suppose we choose an x0 and look at how the displacement varies at this point as t changes. This is the equation of a sinusoid with frequency = u(x,t)=X(x)T(t)=sin nπx L bsin anπt L +ccos anπt L u(x0,t)=Kbsin anπt L +ccos anπt L an 2 L K=sin nπx0 L where n = 1, 2, 3…
- 11. 11 Take it higher So there are three ways to increase the frequency of a sound, producing a “higher” note: frequency= an 2L Increase a (continuous changes) Tuning Increase n (discrete changes) Overblowing Playing harmonics Decrease L (continuous changes) Shortening Fretting
- 12. 12 Boundary conditions With wind instruments, it’s not always true that if L is the length of the tube, u(0, t) = u(L, t) = 0. This is true with the flute (because the pressure doesn’t change at the ends). However, what happens with some instruments (like the sax) is that ux(0, t) = ux(L, t) = 0, which means that the X(x) functions are cosine terms rather than sines. Still more bizarre is the behavior of the clarinet. It has boundary conditions u(0,t)=0 ux(L,t)=0
- 13. 13 Standing waves The harmonics we hear on a stringed instrument and the overtones on a wind instrument are actually produced by standing waves, which are solutions of the form L tan L xn b ππ cossin
- 14. Example :- of one dimentional eqn • Find the solution of the wave equation d2 y/dt2 = c2 d2 y/dx2 such that y=a cos pt when x=l, and y=0 when x=0. 14
- 15. 15 Thank you