====== 10. Waves and sound ======
**--- [[.:10:problems|P R O B L E M S]] ---**

===== - Wave on a stretched string =====
If you pluck a guitar string under tension, a transverse wave moves in the $+x$ direction as shown below.

{{ :courses:phy101:10.1.png?nolink&350 |}}

The net force acting at a particular point on the string (e. g. $x_1,y_1$) is the vector sum of the tension in the string $F_1$ and the restoring force $F_T$. The $x$-components of the two forces cancel out and the sum is only the sum of the $y$-components.

Now, $\tan\theta_1=-F_1/F_T$ is equal to the slope of the function $y(x,t)$ at this point and, hence, the partial derivative of $y$ with respect to $x$, i. e. $\partial y/\partial x$, so

$$ \frac{F_1}{F_T} = - \left(\frac{\partial y}{\partial x}\right)_{x_1}, \ \frac{F_2}{F_T} = \left(\frac{\partial y}{\partial x}\right)_{x_2}. $$

So the net force acting on the small mass element $\Delta m$ with length $\Delta x$ can be written following Newton as

$$ F = F_1 + F_2 = F_T \left[ \left(\frac{\partial y}{\partial x}\right)_{x_2} - \left(\frac{\partial y}{\partial x}\right)_{x_1} \right] = \Delta m a = \mu \Delta x \frac{\partial^2 y}{\partial t^2} $$

because **linear density** of the string $\mu = \Delta m / \Delta x$. Divide both sides by $F_T \Delta x$ and take the limit as $\Delta x$ approaches zero to get

$$ \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \left(\frac{\partial y}{\partial x}\right)_{x_2} - \left(\frac{\partial y}{\partial x}\right)_{x_1} \right] = \frac{\partial^2 y}{\partial x^2} = \frac{\mu}{F_T} \frac{\partial^2 y}{\partial t^2}. $$

Replacing the ration of the curvature and acceleration using the wave equation, we easily find

\begin{equation}
v = \sqrt{\frac{F_T}{\mu}}.
\end{equation}

The speed of a wave on a stretched string only depends on the tension force and the linear density of the string.

===== - Interference and superposition =====
Consider two simple pulses of the same amplitude moving toward one another in the same medium, as shown in Figure  16.6.3 . Eventually, the waves overlap, producing a wave that has twice the amplitude, and then continue on unaffected by the encounter. The pulses are said to interfere, and this phenomenon is known as **interference**.

{{ :courses:phy101:10.2.png?nolink&200 |}}

The **principle of superposition** states that if two or more traveling waves combine at the same point, the resulting position of the mass element of the medium, at that point, is the algebraic sum of the position due to the individual waves. Waves that obey the superposition principle are **linear waves**; they follow the **linear wave equation**. Waves that do not obey the superposition principle are said to be **nonlinear waves**. Waves on a string, sound, surface water waves and electromagnetic waves are linear waves.

Principle of superposition can be used to understand interference. Consider the following wave, instead of just a pulse, to see this.

{{ :courses:phy101:10.3.png?nolink&450 |}}

When two waves (a) and (b) arrive at the same point exactly in phase, their superposition causes like to be added with like. This gives a **constructive interference**.

{{ :courses:phy101:10.4.png?nolink&450 |}}

When two waves (a) and (b) arrive at the same point exactly out of phase, their superposition causes like to be canceled by like. This gives a **destructive interference** and the resultant is zero.

The addition and subtraction of the amplitudes of the two waves can be seen more clearly in the following figure.

{{ :courses:phy101:10.5.png?nolink&500 |}}

Waves $y_1$ and $y_2$ are interfering and the resultant wave $y$ has an amplitude at each point that is the sum of the amplitudes of $y_1$ and $y_2$. This is more clearly in the enlarged part of the wave shown below the first figure. Superposition is exactly this summation of amplitudes.

A general equation for the resultant wave $y$ after an interference of $y_1=A\sin(kx-\omega t + \phi)$ and $y_2=A\sin(kx-\omega t)$ can be derived. According to the principle of superposition,

$$ y = y_1 + y_2 = A[\sin(kx-\omega t + \phi)+\sin(kx-\omega t)] $$

Remembering the trigonometric identity $\sin u + \sin v = 2[\sin(u+v)/2][cos(u-v)/2]$, it can be shown that

\begin{equation} \label{int}
y = 2A \cos\left(\frac{\phi}{2}\right) \sin\left(kx-\omega t + \frac{\phi}{2}\right).
\end{equation}

===== - Standing waves =====
A wave can be reflected from boundaries.

{{ :courses:phy101:10.6.png?nolink&500 |}}

If the boundary is fixed, the reflected wave is exactly $\pi$ radians (180 deg) out of phase with respect to the incident wave. If the boundary is free, the reflected wave is in phase with the incident wave.

The two waves here are then traveling in opposite directions which can give rise to a standing wave.

{{ :courses:phy101:10.7.png?nolink&600 |}}

The red wave $y_1=A\sin(kx-\omega t)$ is traveling in the $+x$ direction and the blue wave $y_2=A\sin(kx+\omega t)$ in the $-x$ direction. They interfere and form the resultant wave

\begin{equation}\label{standing}
y = A[\sin(kx-\omega t)+\sin(kx+\omega t)] = 2A \sin(kx) \cos(\omega t).
\end{equation}

In this derivation, we have used the trigonometric identity $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta \pm \cos\alpha\sin\beta$ where $\alpha=kx$ and $\beta=\omega t$.

In the final expression of $y$, the sine wave is only a function of position and the cosine wave is only a function of time.

The waves are in phase when time is integer multiples of $T/2$. They are out of phase when time is odd multiples of $T/4$. However, at some $x$ positions $y=0$ no matter what the phase relationship between the two waves is. These positions are called **nodes**.

Nodes are the points where $sin(kx)=0$. So $kx$ has to be integer multiples of $\pi$. But $kx=2\pi x/\lambda$. So the nodes are the points where $x$ is an integer multiple of $\lambda/2$.

At the nodes, particles do not oscillate at all. At the **antinodes**, particles oscillate between $y=\pm A$. At the antinodes $\sin(kx)=\pm 1$. So $kx$ has to be odd multiples of $\pi/2$ and, hence, the antinodes are the points where $x$ is an odd multiple of $\lambda/4$.

The resulting black wave in the previous figure is shown more clearly below.

{{ :courses:phy101:10.9.png?nolink&450 |}}

The nodes and antinodes are shown using red and blue dots, respectively.

==== Resonance ====
If a string is fixed at both end and the left end of the string is oscillated vertically using a string vibrator, a wave propagate to the right until getting reflected from the right end. The reflected wave will interfere with incident wave and a standing wave is created only if the interference is constructive.

The interference is constructive only at certain frequencies of the string vibrator called **resonant frequencies**. And the phenomenon of getting a boosted (higher amplitude) standing wave from an initial lower-amplitude wave at certain frequencies is called **resonance**.

{{ :courses:phy101:10.14.png?nolink&550 |}}

The string of length L has a node on either end. There are multiple resonant frequencies. In the figure, four different **modes** of the string is shown numbered $n=1,2,3,4$.

The frequency that causes the first mode (fundamental mode or first harmonic) is called the **fundamental frequency** which is $f_1=v/\lambda_1=v/(2L)$. All frequencies above $f_1$ are called **overtones**.

The second mode or harmonic has wavelength $\lambda=L$ and so on. General equations for the wavelength and frequency of any harmonic or mode ($n$) are

\begin{align}\label{harmonics}
\lambda_n = \frac{2}{n}L \text{ where } n=1,2,3,4,5... \\
f_n = n\frac{v}{2L} \text{ where } n=1,2,3,4,5...
\end{align}

The standing wave patterns shown above are called **normal modes** of a string.

===== - Sound =====
Sound waves propagate through air. In the following figure, a speaker is vibrating in a SHM and the surrounding air is carrying longitudinal sound waves as result.

{{ :courses:phy101:10.8.png?nolink&650 |}}

Patterns of compression and rarefaction give rise to the wave. The wave function can be written in terms of the **gauge pressure** $\Delta P$, the pressure difference with respect to the atmospheric pressure.

$$ \Delta P = \Delta P_{max} \sin(kx \mp \omega t + \phi). $$

Sound wave can also be modeled in terms of the displacement of the air molecules with respect to their equilibrium positions.

$$ s(x,t) = s_{max} \cos(kx \mp \omega t + \phi). $$

These two functions are plotted above for a sound created by a speaker. The pressure and displacement are plotted as a function of distance from the speaker.
===== - Speed and intensity =====
Speed of sound is different in different media. Kinematically the speed depends on the acceleration of the particles or disturbance in the medium $a_y$ and the curvature of the wave $\mathcal{C}$ because the wave equation tells us that the speed of a wave

$$ v = \sqrt{\frac{\text{acceleration of particles}}{\text{curvature of wave}}}. $$

If you recall the relation $\omega=\sqrt{k/m}$, you will see that frequency or velocity of an oscillation depends on the elastic ($k$) and inertial ($m$) properties of a medium. We can generalize this for any medium and also write speed of sound in a medium

$$ v = \sqrt{\frac{\text{elastic property}}{\text{inertial property}}}. $$

If the medium is more rigid or stiff, the parameter describing its elastic property (spring constant $k$ in case of SHM) will be higher and, consequently, speed of sound in that medium will be higher. Conversely, if the density ($\rho$) of the medium is higher, its inertial property is higher and speed of sound in that medium is lower.

In a fluid, speed of sound $v=\sqrt{B/\rho}$ where $B$ is the [[en:bulk-modulus|bulk modulus]] of the fluid.

In a solid, speed of sound $v=\sqrt{Y/\rho}$ where $Y$ is [[en:youngs-modulus|Young's modulus]] of the solid.

In an ideal gas, it can be shown that speed of sound $v=\sqrt{\gamma R T/M}$ where heat capacity ratio $\gamma=C_p/C_V$, $R$ is the universal gas constant, $T$ is the temperature in K and $M$ the molecular mass.

Speed of sound depends on the density of a medium and the density depends on temperature. So speed of sound depends on temperature as well. In air at sea level speed of sound

$$ v = 331 \sqrt{\frac{T}{273}}. $$

Intensity of sound waves decreases with distance. Sound waves propagate in all directions. If the power of a sound wave is $P$ and the distance to source of the wave is $r$, then sound **intensity**

$$ I = \frac{P}{4\pi r^2}.$$

This is true because $4\pi r^2$ is the area of the surface of a sphere of radius $r$. Here, usually the time averaged power $<P>$ is used. We can use this relation to derive an expression for intensity of sound wave in air, symbolically

$$ I = \frac{(\Delta p_{max})^2}{2\rho v} $$

where $\Delta p_{max}$ is the maximum gauge pressure, $\rho$ density of air and $v$ speed of sound in air.

We usually use **sound intensity level** which is a relative quantity. It is the intensity of sound with respect to the a reference intensity. The reference is the threshold intensity of hearing at 1000 Hz, written as $I_0=10^{-12}$ W m$^{-2}$. All other intensities are measured with respect to $I_0$ in **decibel** (dB) units. The sound intensity level (in dB)

$$ \beta = 10 \log_{10} \frac{I}{I_0}. $$

Intensity in an average university classroom is around $10^{-4}$ W m$^{-2}$, so the intensity level is $80$ dB.

===== - Sound interference =====
If two waves with different phase shifts interfere with each other, the resultant wave is described by Eq. \eqref{int}. The phase shift can occur if two wave travel different distances, if they have different path lengths. In the following figure, two sound waves are created from two speakers and the waves travel different path lengths $r_1$ and $r_2$ to reach the point $P$.

{{ :courses:phy101:10.10.png?nolink&500 |}}

The difference in the path lengths of the two waves when they reach point $P$ is $\Delta r = |r_2-r_1|$.

When $\Delta r$ is an integer multiple of $\lambda$, the waves are in phase and we get constructive interference.

When $\Delta r$ is an odd multiple of $\lambda/2$, the waves are 180 deg out of phase and we get destructive interference.

===== - Standing sound waves and music =====
**Standing wave** is created when two waves traveling in opposite directions interfere. **Resonance** occurs when higher amplitude sound is produced through constructive interference in such a wave. We can have standing wave and resonance for sound waves.

{{ :courses:phy101:10.11.png?nolink&300 |}}

Sound from a tuning fork is hitting the left end of a tube open to the right. The wave is reflected and standing waves are created for certain frequencies of the fork. There is a node at closed end and an antinode at the open end if the wavelength of a wave from the fork $\lambda_1=4L$ because the distance between a node and antinode is one-fourth of a wavelength.

{{ :courses:phy101:10.12.png?nolink&700 |}}

This is the first or fundamental frequency, but there are overtones as well. The first three overtones are shown above. In general for a tube closed at one end resonant wavelengths $\lambda_n=4L/n$ and the resonant frequencies $f_n=nv/(4L)$. Here $n$ must be an odd number.

{{ :courses:phy101:10.13.png?nolink&700 |}}

For a tube open at both ends, there must be an antinode at either end. This condition directly gives the resonance conditions. The resonant wavelengths and frequencies are given by

\begin{align}\label{tube}
\lambda_n = \frac{2}{n}L \text{ where } n=1,2,3,4... \\
f_n = n\frac{v}{2L} \text{ where } n=1,2,3,4...
\end{align}

These are the wavelengths and frequencies of the normal modes of a standing sound wave in hollow tube open at both ends, like a flute.
===== - Doppler effect =====
Relative motion of source (of a wave) and observer (of the wave) toward one another increases the received frequency. Relative motion apart decreases frequency. The greater the relative speed, the greater the effect. This is called **Doppler effect**.

{{ :courses:phy101:10.15.png?nolink&650 |}}

If the source of a sound wave (the car) is moving toward the observer Y, successive emissions occurs at closer distances and, hence, the waves are squeezed, wavelength received by Y decreases, frequency increases. The opposite happens for the observer X. If, on the other hand, the source is stationary and the observers are moving, X will receive the waves at lower frequency and Y at higher frequency.

If $f_s$ is frequency at which the source emits, $f_o$ the frequency at which observer receives, $v_s$ and $v_o$ are the velocities of the source and observer, respectively, and $v$ is the speed of the wave, then

\begin{equation}
f_o = f_s \frac{v\pm v_o}{v\mp v_s}.
\end{equation}

The top sign is for approaching and the bottom sign is for departing. Let us derive the case for a source moving away from observer $X$ and moving toward $Y$.

{{ :courses:phy101:10.17.png?nolink&700 |}}

Snapshots of the source at an interval of $T_s$ are shown as it moves away from X with speed $v_s$. The solid lines indicate the actual position of the wavefronts and the dotted line the positions if the source was not moving. Now received wavelength by observer X is $\lambda_o = \lambda_s + \Delta x 
\Rightarrow vT_0=vT_s+v_sT_s$ which gives

$$ \frac{v}{f_o} = \frac{v}{f_s} + \frac{v_s}{f_s} \Rightarrow f_o = f_s\left(\frac{v}{v+v_s}\right). $$

For the observer Y we can begin from $\lambda_o = \lambda_s - \Delta x$ and come to the conclusion that

$$ f_o = f_s\left(\frac{v}{v-v_s}\right). $$