PHYS212 : Lectures

Subsections

• 1.5.1 The Equation of Motion : The Wave Equation
• 1.5.2 Solutions : Plane Waves
• 1.5.3 Normal Modes : Standing Waves

1.5 The Continuous Limit : Waves on a String

1.5.1 The Equation of Motion : The Wave Equation

• Eq. 1.36 can be re-written
  

  $$\ddot{y_b} = -\frac{T_o}{m} \left( \frac{y_b - y_{b-1}}{\ell} - \frac{y_{b+1} - y_b}{\ell} \right).$$ (1.51)

  
  

• Noting that the horizontal distance $\Delta x$ between the beads is approximately $\ell$, we can re-frame Eq. 1.51 as a finite difference equation,
  

  $$\ddot{y_b} = \frac{T_o}{m} \frac{[(y_{b+1} - y_b) - (y_b - y_{b-1})]}{\Delta x}$$

  $$= \frac{T_o}{m} \frac{\Delta(\Delta y)}{\Delta x}$$

  $$= \frac{T_o \ell}{m} \frac{\Delta(\Delta y)}{(\Delta x)^2}$$ (1.52)

  
  

• Taking the limit as the separation between masses $\Delta x$ goes to zero, and defining the linear mass density of the system as
  

  $$\mu = \frac{m}{\ell},$$ (1.53)

  
  
  we obtain the equation of motion for the transverse motion of a stretched string
  

  $$\ddot{y} = \frac{T_o}{\mu} y''$$ (1.54)

  
  
  where $y'' = \frac{\partial^2 y}{\partial x^2}$. This is also called the wave equation.

1.5.2 Solutions : Plane Waves

• Separation of Variables
  • Eq. 1.55 is a linear differential equation in two independent variables $x$ and $t$. We can split equations of this kind into two equations, each having one independent variable, using the aptly-named method of separation of variables.

  • We first propose a product solution
    

    $$y(x,t) = f(x) g(t)$$ (1.55)

    
    
    which we can use to rewrite Eq. 1.55 as
    

    $$f(x) \ddot{g}(t) = \frac{T_o}{\mu} g(t) f''(x)$$ (1.56)

    
    

  • Dividing both sides of Eq. 1.57 by the product solution $f(x) g(t)$ yields
    

    $$\frac{\ddot{g}(t)}{g(t)} = \frac{T_o}{\mu} \frac{f''(x)}{f(x)}.$$ (1.57)

    
    

  • The left side of Eq. 1.58 is dependent only on $t$, and the right side is dependent only on $x$. In order for the equation to hold for all values of the independent variables $x$ and $t$, both sides must be equal to the same constant. Recognizing the time-dependent equation, we call this constant $-\omega^2$, and we have the two equations,
    

    $$\ddot{g} = -\omega^2 g$$

    $$f'' = -\frac{\omega^2 \mu}{T_o} f = -k^2 f.$$ (1.58)

    
    

  • The solutions to both equations are sinusoidal functions or complex exponentials,
    

    $$f(x) = \left\{ \begin{array}{l} A \cos(kx) + B \sin(kx)\\ C Re\left\{ e^{i(kx + \phi_x)} \right\} \end{array}\right.$$

    $$g(t) = \left\{ \begin{array}{l} A \cos(\omega t) + B \sin(\omega t)\\ C Re\left\{ e^{i(\omega t + \phi_t)} \right\} \end{array}\right.$$ (1.59)

    
    

  • By convention, the spatial angular frequency is called the wavenumber and is represented by $k$. It has units of radians per unit length and is related to the wavelength via
    

    $$k = \frac{2\pi}{\lambda}$$ (1.60)

    
    
    It should not be confused with a spring constant.

  • It follows from Eq. 1.59 that waves on a string have
    

    $$k = \sqrt{\frac{\mu}{T_o}} \omega.$$ (1.61)

    
    

• Plane Waves
  • You will show for homework, via the sinusoidal solutions, that the separate position-dependent and time-dependent solutions of Eq.'s 1.60 can be combined to give solutions taking the forms
    

    $$y(x,t) = \left\{ \begin{array}{l} A \cos(kx -\omega t) + B ... ...left\{ e^{i(kx - \omega t + \phi)} \right\} \end{array}\right.$$ (1.62)

    
    

  • These solutions are called plane waves, because, in three dimensions, they describe wave fronts of equal displacement in the $\hat{\jmath}$-$\hat{k}$ plane.

  • General solutions of the wave equation consist of linear combinations of plane waves of different $k$.

  • Animation : planeWave.py (see Appendix A on running animations)

    Shows a sample wave in time and space.

  • Animation : plusOrMinus.py (see Appendix A on running animations)
    Note : In order for this animation to run, the file planeWave.py must be present in the same directory.

    An answer to the question : Why did we choose $(kx -\omega t + \phi)$ instead of $(kx + \omega t + \phi)$ for the phase of our solutions in Eq. 1.63?

    Note that the animation shows two ways (each) to produce plane waves traveling in $\hat{\imath}$ and $-\hat{\imath}$.

• Phase Velocity
  • Consider the point on a plane wave corresponding to a value of zero phase,
    

    $$kx - \omega t + \phi = 0$$ (1.63)

    
    

  • Solving for $x$, we find that as time passes, this point on the wave form follows a trajectory along $\hat{\imath}$ given by
    

    $$x = \frac{\omega}{k} t - \phi.$$ (1.64)

    
    

  • Hence, we conclude that the wave moves with a constant velocity
    

    $$v_p = \frac{\omega}{k}$$ (1.65)

    
    
    called the phase velocity.

  • The plane wave solutions to the wave equation (Eq. 1.63) can be expressed in terms of the phase velocity as
    

    $$y(x,t) = \left\{ \begin{array}{l} A \cos[k(x - v_p t)... ...ft\{ e^{i[k(x - v_p t) + \phi]} \right\} \end{array} \right.$$ (1.66)

    
    

  • It follows from Eq.'s 1.62 and 1.66 that waves on a string have phase velocities of magnitude
    

    $$v_p = \sqrt{\frac{T_o}{\mu}}$$ (1.67)

    
    

• The General Form of the Wave Equation
  • We will encounter the wave equation in many contexts beyond stretched strings.

  • A more general form is
    

    $$\ddot{y} = \frac{\omega^2}{k^2} y'' = v_p^2 y''$$ (1.68)

    
    

1.5.3 Normal Modes : Standing Waves

• A string stretched between two stationary points a distance $L$ apart is the continuous limit of the beaded string we considered in Section 1.4.1.

• In the continuous limit, the number of beads has become effectively infinite, and we expect an infinite set of normal modes, called standing waves, satisfying the boundary conditions $y(0,t) = y(L,t) = 0$.

• The condition $y(0,t) = 0$ is automatically satisfied if the spatial part of the solutions take the form $A \sin(kx)$.

• With these solutions, it follows from $y(L,t)$ = 0 that
  
  $$n \frac{\lambda}{2} = L \hspace{48pt} (n = 1, 2, 3, ...)$$
  
  
  which means
  

  $$k_n = \frac{n \pi}{L} \hspace{48pt} (n = 1, 2, 3, ...)$$ (1.69)

  
  
  and
  

  $$\omega_n = \sqrt{\frac{T_o}{\mu}} k_n = \sqrt{\frac{T_o}{\mu}} \frac{n \pi}{L} \hspace{48pt} (n = 1, 2, 3, ...)$$ (1.70)

  
  

• The alternative form of the sine function
  

  $$\sin(kx) = \frac{\left( e^{ikx} - e^{-ikx} \right)}{2i} = \frac{\left( e^{ikx} + e^{i(-kx + \pi)} \right)}{2i}$$ (1.71)

  
  
  indicates that we are looking for combinations of two plane waves with wavenumbers $k_n$ and $-k_n$ with a relative phase of $\phi = \pi$ (or a relative minus sign),
  

  $$y(x,t) = Re\left\{ A\left( e^{i(k_nx - \omega_n t)} - e^{i(-k_nx - \omega_n t)} \right) \right\}$$ (1.72)

  
  
  Hence, standing waves consist of a superposition of plane waves of equal amplitude traveling in opposite directions.

• Animation : standingWave.py (see Appendix A on running animations)
  Note : In order for this animation to run, the file planeWave.py must be present in the same directory.

• If the constant $A$ is imaginary Eq. 1.73 is equivalent to
  

  $$y(x,t) = Re\left\{ A\left( e^{ik_nx} - e^{-ik_nx} \right) e^{-i\omega_nt} \right\}$$

  $$= Re\left\{ 2iA \sin(k_nx) e^{-i\omega_nt} \right\}$$

  $$= -2iA \sin (k_nx) \cos(\omega_n t)$$ (1.73)

  
  
  and if $A$ is real, we have
  

  $$y_n(x,t) = 2A \sin (k_nx) \sin(\omega_n t)$$ (1.74)

  
  
  These describe standing waves satisfying different initial conditions.

• General standing wave solutions are
  

  $$y_n(x,t) = \left\{ \begin{array}{l} \sin(k_nx) \left[ A\cos(... ...in(k_nx) e^{-i(\omega_n t - \phi)} \right\} \end{array}\right.$$ (1.75)

  
  

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006