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Chapter 17 - Traveling Waves

Pulses

Suppose we consider a 1 dimensional medium to be a number of masses connected to each other by springs.

The Shive Wave Machine is essentially such a system, the rods in the machine are connected to each other by a wire (you can think of each element as torsional pendulum. When one spine is displaced the next one feels a force pushing in the same direction, if a continuous force is applied to the system the whole system can move to a new equilibrium.

However if a short disturbance is applied a pulse is observed. When the force at the source is removed the rods are driven back to the their equilibrium position by the restoring force, just like in simple harmonic motion. The duration of the pulse on the machine depends on the time the driving force is applied.

We can see that the velocity of the rods is in a different direction and of a different magnitude to the velocity of the pulse.

From pulses to waves

The displacement in a pulse can be either perpendicular to, or parallel to the direction of motion, resulting in either transverse or longitudinal pulses.

If the force at the source is changed from a transient one to a continuous oscillatory one then the result is a wave. If we freeze the wave at a certain time the displacement of the points can often be represented a sinusoidal function, $D(x)=A\sin\frac{2\pi}{\lambda}x$.

If the wave moves so that it takes a time $T$, the period, for a wavelength $\lambda$ to pass a point we can say that the velocity of a wave is $v=\frac{\lambda}{T}=f\lambda$.

Traveling waves

In order to have a wave which at time $t=0$ has a displacement function $D(x)=A\sin\frac{2\pi}{\lambda}x$ propagate with $v$ we can write

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

which using $v=\frac{\lambda}{T}=f\lambda$ can be written

$D(x,t)=A\sin(\frac{2\pi}{\lambda}x-\frac{2\pi t}{T})=A\sin(kx-\omega t)$

We have here defined a new quantity $k$ which is the wave number, $k=\frac{2\pi}{\lambda}$ and expressed the frequency in terms of the angular frequency $\omega=2\pi f$

The velocity of the waves propagation, which we call the phase velocity can be expressed in terms of $\omega$ and $k$

$v=f\lambda=\frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$

17.P.012

Wave equation

The equation for a wave's motion we have just considered

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

is a solution to the a differential equation that we call the wave equation.

$v^2\frac{\partial^{2} D}{\partial x^2}=\frac{\partial^{2} D}{\partial t^2}$

We can check that this is the case by substituting in our equation of motion in to the differential equation

$-v^2(\frac{2\pi}{\lambda})^{2}A\sin\frac{2\pi}{\lambda}(x-vt)=-v^{2}(\frac{2\pi}{\lambda})^{2}A\sin\frac{2\pi}{\lambda}(x-vt)$

We can now look for physical situations that give rise the wave equation, these will define the values of the velocity in terms of the properties of the medium through which the wave propagates.

Waves on a string

We define a piece of the string which has length $\Delta x$, we assume that the amplitude of the wave is fairly small so that the angles $\alpha$ and $\beta$ are small. We then set the two forces equal in magnitude $F_{1}=F_{2}=F_{T}$

The vertical force on the element of the string is then

$-F_{T}\sin\beta-F_{T}\sin\alpha$

From Newton's second law

$-F_{T}\sin\beta-F_{T}\sin\alpha=ma_{y}=\mu\Delta x \frac{\partial^{2} D}{\partial t^2}$

where $\mu$ is the linear mass density of the string ($\mu=\frac{m_{string}}{l_{string}}$)

As the angles $\alpha$ and $\beta$ are small we can say that

$\sin\alpha\approx\tan\alpha=\frac{\partial D}{\partial x}|^{x}$ and $\sin\beta\approx\tan\beta=-\frac{\partial D}{\partial x}|^{x+\Delta x}$

We can the rearrange our equation to be $\Large \frac{(\frac{\partial D}{\partial x}|^{x+\Delta x}-\frac{\partial D}{\partial x}|^{x})}{\Delta x}=\frac{\mu}{F_{T}}\frac{\partial^{2} D}{\partial t^2}$

Taking the limit $\Delta x \to 0$ we get $\large \frac{\partial^{2} D}{\partial x^2}=\frac{\mu}{F_{T}}\frac{\partial^{2} D}{\partial t^2}$

Which is the wave equation with $v=\sqrt{\frac{F_{T}}{\mu}}$

17.P.027

Pressure waves

Sound can be represented as a longitudinal wave

$D=A\sin(kx-\omega t)$

The change in pressure from the background pressure $P_{0}$ in response to a volume change is related to the bulk modulus $B$

$\Delta P=-B\frac{\Delta V}{V}=-B\frac{A(D_{2}-D_{1})}{A(x_{2}-x_{1})}$ which in the limit of $\Delta x \to 0$ is $\Delta P=-B\frac{\partial D}{\partial x}$

$\Delta P=-BAk\cos(kx-\omega t)$

Displacement and pressure

$D=A\sin(kx-\omega t)$

$\Delta P=-BAk\cos(kx-\omega t)$

$\Delta P$ is the sound pressure level, the deviation from the background pressure.

Speed of Sound

We now look at the speed of a longitudinal wave, such as sound, from the perspective of the wave equation. The change in pressure from the background pressure $P_{0}$ in response to a volume change is related to the bulk modulus $B$

$\Delta P=-B\frac{\Delta V}{V}=-B\frac{A(D_{2}-D_{1})}{A(x_{2}-x_{1})}$ which in the limit of $\Delta x \to 0$ is $\Delta P=-B\frac{\partial D}{\partial x}$

The force on the element is given by the difference in the pressure

$F=A(P_{1}-P_{2})=A(\Delta P_{1}-\Delta P_{2})$

Newton's Second Law gives us that

$A(\Delta P_{1}-\Delta P_{2})=m\frac{\partial^{2}D}{\partial t^{2}}=\rho A(x_{2}-x_{1})\frac{\partial^{2}D}{\partial t^{2}}$ → $\frac{\Delta P_{2}-\Delta P_{1}}{x_{2}-x_{1}}=-\rho\frac{\partial^{2}D}{\partial t^{2}}$

which in the limit $\Delta x \to 0$ is

$\frac{\partial \Delta P}{\partial x}=-\rho\frac{\partial^{2}D}{\partial t^{2}}$

and using our result connecting the pressure to the bulk modulus

$-B\frac{\partial^{2}D}{\partial x^{2}}=-\rho\frac{\partial^{2}D}{\partial t^{2}}$

which is the wave equation with $v=\sqrt{\frac{B}{\rho}}$

The speed of sound in air depends on temperature according to

$v=331\sqrt{1+\frac{T_{C}}{273^{\circ} C}}\mathrm{ms^{-1}}$

where $T$ is temperature in $\mathrm{^{\circ}C}$.

Energy in a wave

Waves transport energy. How much?

We can consider each particle in a wave to be undergoing simple harmonic motion with total energy given by $E=\frac{1}{2}kA^{2}$ where $k$ is the effective spring constant and $A$ is the amplitude of the wave. The frequency equation for SHM $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ tell us that $k=4\pi^{2}mf^{2}$ and we can rewrite the energy as

$E=2\pi^{2}mf^{2}A^{2}$

The mass in this equation can be written as $m=\rho V$, where $\rho$ is the density of the medium and $V$ is the volume through which the wave passes. If we are looking at 3 dimensional wave that traveling outwards from a point, we can say it travels a distance $l=vt$ in a time interval $t$. This lets us define our volume as $V=Sl$ where S is the cross-sectional area through which the wave passes. Combining all this, $m=\rho V=\rho Sl=\rho Svt$ and the energy transported through a surface $S$ in time $t$ is

$E=2\pi^{2}\rho Svtf^{2}A^{2}$

and the average power

$\bar{P}=\frac{E}{t}=2\pi^{2}\rho Svf^{2}A^{2}$

The intensity is the average power per unit area

$I=\frac{\bar{P}}{S}=2\pi^{2}\rho vf^{2}A^{2}$

As a 3 dimensional wave propagates from a point the area through which the wave passes increases. When the power output is constant, the intensity decreases as $\frac{1}{r^{2}}$, so the ratio of the intensity at $r_{1}$, $I_{1}$, to the intensity at $r_{2}$, $I_{2}$ is

$\frac{I_{2}}{I_{1}}=\frac{\bar{P}/4\pi r_{2}^{2}}{\bar{P}/4\pi r_{1}^{2}}=\frac{r_{1}^2}{r_{2}^{2}}$

In order for the power output to be constant the amplitude must also decrease with $S_{1}A_{1}^2=S_{2}A_{2}^2$ implying $A\propto\frac{1}{r}$

17.P.043

Loudness and decibels

Our sensitivity to the loudness of sound is logarithmic, a sound that is ten time as intense sounds only twice as loud to us. The sound level $\beta$ is thus measured on a logarithmic scale in decibels is

$\beta=10\log_{10}\frac{I}{I_{0}}$

$I_{0}$ is the weakest sound intensity we can hear $I_{0}=1.0\times 10^{-12}\mathrm{W/m^{2}}$

Some examples of different sounds loudness in decibels can be found here.

Our hearing is not equally sensitive to all frequencies, you can test your hearing here

17.P.046

Refraction and Diffraction

Here we will only look at refraction and diffraction briefly (we will take a look at them in the ripple tank!), they will be covered in more detail in PHY132 when we look at optics.

Refraction occurs when 2 or 3 dimensional waves pass into a medium where they have a different velocity, when they do this they change their direction.

$\frac{\sin\theta_{2}}{\sin\theta_{1}}=\frac{v_{2}}{v_{1}}$

We can see extensive diffraction in the ripple tank around various objects and we can note that the angular spread of waves $\theta$ around an object is

$\theta\approx\frac{\lambda}{l}$

where $\lambda$ is the wavelength of the waves and $l$ is the width of the opening or object about which diffraction occurs.

17.P.051

The Doppler effect

Doppler Effect with moving source

When a source of sound moves a stationary observer hears an apparent shift in the frequency. The origin of this effect can be seen nicely in this animation from Wikipedia.

For a source moving towards an observer

$\lambda'=\lambda-d=\lambda-v_{source}T=\lambda-v_{source}\frac{\lambda}{v_{sound}}=\lambda(1-\frac{v_{source}}{v_{sound}})$

$f'=\frac{v_{sound}}{\lambda'}=\frac{v_{sound}}{\lambda(1-\frac{v_{source}}{v_{sound}})}=\frac{f}{(1-\frac{v_{source}}{v_{sound}})}$

If the source is moving away from the observer

$\lambda'=\lambda(1+\frac{v_{source}}{v_{sound}})$

$f'=\frac{f}{(1+\frac{v_{source}}{v_{sound}})}$

Doppler effect with moving observer

A doppler effect also occurs when an observer moves towards a source, but here the wavelength does not change, instead it is the effective velocity that changes and leads to an apparent change in the frequency of the sound. When the observer moves towards the source of the sound

$T'=\frac{\lambda}{v_{sound}+v_{obs}}$

$f'=\frac{v_{sound}+v_{obs}}{\lambda}=\frac{v_{sound}+v_{obs}}{v_{sound}}f$

When the observer moves away from the source of the sound

$f'=\frac{v_{sound}-v_{obs}}{\lambda}=\frac{v_{sound}-v_{obs}}{v_{sound}}f$

Doppler Effect with moving observer and source

For an observer moving towards a source which is also moving towards it

$f'=\frac{v_{sound}+v_{obs}}{\lambda'}=\frac{v_{sound}+v_{obs}}{\lambda(1-\frac{v_{source}}{v_{sound}})}=\frac{(v_{sound}+v_{obs})v_{sound}}{\lambda(v_{sound}-v_{source})}=f\frac{(v_{sound}+v_{obs})}{(v_{sound}-v_{source})}$

A general formula for the doppler effect is

$f'=f\frac{(v_{sound}\pm v_{obs})}{(v_{sound}\mp v_{source})}$

Top part of the $\pm$ or $\mp$ sign is for a source or observer moving towards each other, the bottom part is for motion away from each other.

17.P.057

Double Doppler effect when sound bounces off a moving object

If we project sound at a moving object and wait for it to come back it's frequency is shifted by two doppler effects. In the first part of the process the source is stationary and the observer is moving, the moving object is hit by sound of frequency $f'$

$f'=f\frac{(v_{sound}\pm v_{obs})}{(v_{sound})}$

Now when the sound is returning to us it is a case of a moving source and stationary observer, we apply the formula for that on to the already shifted frequency $f'$ to get $f''$ the frequency we would hear

$f''=f'\frac{(v_{sound})}{(v_{sound}\mp v_{source})}=f\frac{(v_{sound})}{(v_{sound}\mp v_{source})}\frac{(v_{sound}\pm v_{obs})}{(v_{sound})}$

Of course in this case $v_{source}=v_{obs}=v_{object}$ so

$f''=f\frac{(v_{sound}\pm v_{object})}{(v_{sound}\mp v_{object})}$

This double doppler effect, used with either sound (sonar) or radio waves (radar) can be used to detect the speed of an object from the frequency of the reflected waves.

Sonic Boom

When a moving object moves faster than the speed of sound the wavefronts pile up, creating a shock wave, which is heard as a sonic boom.

A video of a sonic boom, and another one.

More on sonic booms on wikipedia.

phy131studiof15/lectures/chapter17.txt · Last modified: 2015/11/03 23:21 by mdawber
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