close

<Derivation of the Classical Wave Equation>

Consider a vibrating string at a instant time:

Wave01.png

Focus on the red segment:

Wave02.gif

Assume that the string is homogeneous, and the mass of the string per unit length is r (a constant).

T1 & T2: tension acting on P & Q, and T1 & T2 are tangential to the curve of string.

Only vertical motion exists, that is, the horizontal components of the tensions at all points along the string must be equal.

Consider the horizontal part:

wave03.gif

Consider the vertical force:

wave04.gif

Upward motion is positive, “+”; Downward motion is negative, “-”.

According to Newton’s 2nd law, F = ma = rDx     wave05.gif

wave06.gif

tanb and tana is the slope of curve at x & x+Dx.

that is,

wave07.gif

wave08.gif

and let

wave09.gif

wave10.gif

v is the speed with which a disturbance propagates along a stretched string depends on the tension in the string and the linear mass density of the string.

 

<How to Solve it & Discussion>

Try separation of variables

y(x,t) = X(x)T(t)

wave11.gif

Multiply 1/y on both side

wave12.gif

x and t are independent variables. This equation can only be valid if both sides are equal to a constant. 

wave13.gif

Note we have total not partial derivatives: linear, 2nd-order, and ordinary differential equation.

General solutions of have the form (see the below tables)

  Possible Solution General Solution  
K > 0 ekx, e-kx X(x) = A ekx + Be-kx K = k2
K < 0 sin kx, cos kx X(x) = C sin kx + D cos kx K = -k2

Also for T(t) equation

wave14.gif

K > 0 T(t) = E evkt + F e-vkt
K < 0 T(t) = G sin vkt + H cos vkt

Now look at Boundary Conditions

Wave15.png

y(0,t) = 0

y(L,t) = 0

For K > 0, try to satisfy boundary conditions

Wave16.gif

so we should study the  condition of〝K < 0 〞

For K < 0

Wave17.gif

=> quantization

Wave18.gif

Wave19.gif

In the same way, we can solve T(t) when T(0) = 0.

Wave20.gif

Wave21.gif

Now we choose the initial condition y = y0, that is J = y0

Wave22.gif

This is a set of solutions, one for each value of n. Each solution corresponds to a different standing wave. The differential equation, the boundary conditions, and the initial conditions have completely determined the set of wave functions.

The velocity is given by

Wave23.gif

The velocity at t = 0

Wave24.gif

Our initial condition has specified the initial velocity. Instead of specifying the maximum amplitude we could have specified that the initial velocity is given by this function.

Typically for a standing wave, its wavelength (l) equals to 2L/n (n: number of standing wave), i.e. l is quantized.

Wave25.gif

The period t of our standing wave is the time required for the argument of the time factor sin(npvt/L) to change by 2p, so that

Wave26.gif

The frequency is quantized. It is directly proportional to the integer n, inversely proportional to the length of the string, directly proportional to the square root of the tension force, and inversely proportional to the square root of the mass per unit length.

Wave27.png

A string does not usually move as described by a single harmonic. A linear combination of harmonics can satisfy the wave equation:

Wave28.gif

The fact that a linear combination of solutions can be a solution to the wave equation is called the principle of superposition.

Wave29.png

 

Reference: 

1. Physical Chemistry 3rd Ed Robert G Mortimer

2. http://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2013/

 

 

 

 

 

 

 

 

 

arrow
arrow

    albert1225 發表在 痞客邦 留言(0) 人氣()