7.4 Transverse Waves on a String

Consider an infinite string under constant tension $T$ . We will now show that the equation of motion of the string is the wave equation and derive the wave speed. Consider a short section of the string of length $\Delta x$ . We are assuming that the string has linear density $\mu$ , zero stiffness, and we are ignoring the effects of gravity. Then assuming that there are only small displacements on the string, then $\frac{\partial y}{\partial x}$ is small, so the angles $\theta_{1}$ and $\theta_{2}$ are also small. Hence we use the small angle approximation and say that $\cos\theta_{1}\approx\cos\theta_{2}\approx 1$ .

	$\displaystyle\sum F_{x}$	$\displaystyle=-\lvert\boldsymbol{T}_{1}\rvert\cos\theta_{1}+\lvert\boldsymbol{% T}_{2}\rvert\cos\theta_{2}=0$		(7.39)
	$\displaystyle\implies\lvert T_{1,x}\rvert$	$\displaystyle\approx\lvert T_{2,x}\rvert\approx T.$		(7.40)

From these we get that $T_{1,x}\approx-T$ and $T_{2,x}\approx T$ .

Now, using some trigonometry, notice that

	$\displaystyle\left.\frac{\partial y}{\partial x}\right\|_{x}$	$\displaystyle=\frac{T_{1,y}}{T_{1,x}}\approx-\frac{T_{1,y}}{T}$		(7.41)
	$\displaystyle\left.\frac{\partial y}{\partial x}\right\|_{x+\Delta x}$	$\displaystyle=\frac{T_{2,y}}{T_{2,y}}\approx\frac{T_{2,y}}{T}.$		(7.42)

Thus the net force in the $y$ -direction is given by

	$\displaystyle F_{y}$	$\displaystyle=T_{1,y}+T_{2,y}$		(7.43)
		$\displaystyle=T\left(-\left.\frac{\partial y}{\partial x}\right\|_{x}+\left.% \frac{\partial y}{\partial x}\right\|_{x+\Delta x}\right).$		(7.44)

Using Newton’s second law, we get

$\displaystyle F_{y}$	$\displaystyle=ma_{y}$	(7.45)
	$\displaystyle=\mu\Delta xa_{y}$	(7.46)
	$\displaystyle=\mu\Delta x\left.\frac{\partial^{2}y}{\partial t^{2}}\right\|_{x+% \frac{\Delta x}{2}}=\left(\left.\frac{\partial y}{\partial x}\right\|_{x+\Delta x% }-\left.\frac{\partial y}{\partial x}\right\|_{x}\right)T.$	(7.47)

Finally we divide by $\Delta x$ on both sides and take the limit as $\Delta x\to 0$ to get

$\displaystyle\lim_{\Delta x\to 0}\left(\mu\left.\frac{\partial^{2}y}{\partial t% ^{2}}\right\|_{x+\frac{\Delta x}{2}}\right)$	$\displaystyle=\lim_{\Delta x\to 0}\left[\frac{\left(\left.\frac{\partial y}{% \partial x}\right\|_{x+\Delta x}-\left.\frac{\partial y}{\partial x}\right\|_{x}% \right)T}{\Delta x}\right]$	(7.48)
$\displaystyle\mu\frac{\partial^{2}y}{\partial t^{2}}$	$\displaystyle=T\frac{\partial^{2}y}{\partial x^{2}}$	(7.49)
$\displaystyle\implies\frac{\partial^{2}y}{\partial t^{2}}$	$\displaystyle=\frac{T}{\mu}\frac{\partial^{2}y}{\partial x^{2}}.$	(7.50)

This is the wave equations and we can see that for waves on a string, $v=\sqrt{\frac{T}{\mu}}$ .

7.4.1 Energy in a Wave

What is the mechanical energy stored in a wave on a string? It will have two contributions, potential energy which depends on the displacement of every point and kinetic energy which depends on the velocity of every point. Consider a segment of the string of length $\mathrm{d}x$ , mass $\mathrm{d}m=\mu\mathrm{d}x$ . The infinitesimal contribution to the kinetic energy of the wave is given by

\mathrm{d}K=\frac{1}{2}\mathrm{d}mv_{y}^{2}=\frac{1}{2}\mu\mathrm{d}x\left(% \frac{\partial y}{\partial t}\right)^{2}.

(7.51)

To get a value for this, we integrate it over some length $L$ .

K=\frac{1}{2}\mu\int_{0}^{L}\left(\frac{\partial y}{\partial t}\right)^{2}% \mathrm{d}x.

(7.52)

The potential energy is due to the stretching of the string. A segment of length $\mathrm{d}x$ stretches to a length $\mathrm{d}s$ , and we can calculate the relationship between the two as follows:

$\displaystyle\mathrm{d}s$	$\displaystyle=\sqrt{\mathrm{d}x^{2}+\mathrm{d}y^{2}}$	(7.53)
	$\displaystyle=\sqrt{\mathrm{d}x^{2}+\mathrm{d}x^{2}\left(\frac{\partial y}{% \partial x}\right)^{2}}$	(7.54)
	$\displaystyle=\mathrm{d}x\sqrt{1+\left(\frac{\partial y}{\partial x}\right)^{2}}$	(7.55)
	$\displaystyle\approx\mathrm{d}x\left(1+\frac{1}{2}\left(\frac{\partial y}{% \partial x}\right)^{2}\right),$	(7.56)

where in the last line we have used the Taylor expansion $\sqrt{1+u^{2}}\approx 1+\frac{1}{2}u^{2}$ when $u$ is small. This means we can calculate the potential energy as

$\displaystyle\mathrm{d}U$	$\displaystyle=T(\mathrm{d}s-\mathrm{d}x)$	(7.57)
	$\displaystyle\approx\frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^{2}% \mathrm{d}x$	(7.58)
$\displaystyle\implies U$	$\displaystyle=\frac{1}{2}T\int_{0}^{L}\left(\frac{\partial y}{\partial x}% \right)^{2}\mathrm{d}x.$	(7.59)

Example 7.5.

Consider a sinusoidal wave $y=A\cos(kx-\omega t)$ . What is the energy per unit wavelength? The partial derivatives are given by

	$\displaystyle\frac{\partial y}{\partial t}$	$\displaystyle=A\omega\sin(kx-\omega t)$		(7.60)
	$\displaystyle\frac{\partial y}{\partial x}$	$\displaystyle=-Ak\sin(kx-\omega t),$		(7.61)

so the infinitesimal contribution to the total energy is

$\displaystyle\mathrm{d}E$	$\displaystyle=\mathrm{d}K+\mathrm{d}U$	(7.62)
	$\displaystyle=\frac{1}{2}\left[\mu\left(\frac{\partial y}{\partial t}\right)^{% 2}+T\left(\frac{\partial y}{\partial x}\right)^{2}\right]\mathrm{d}x$	(7.63)
	$\displaystyle=\frac{1}{2}A^{2}\sin^{2}(kx-\omega t)(\mu\omega^{2}+Tk^{2})% \mathrm{d}x.$	(7.64)

Note that $v=\frac{\omega}{k}=\sqrt{T}{\mu}$ , so $Tk^{2}=\mu\omega^{2}$ . Hence for a sinudoidal wave, the kinetic and potential energies are the same. The energy per unit wavelength is then

	$\displaystyle E_{\lambda}$	$\displaystyle=\mu A^{2}\omega^{2}\int_{0}^{\lambda}\sin^{2}(kx)\mathrm{d}x$		(7.65)
		$\displaystyle=\frac{1}{2}\lambda\mu A^{2}\omega^{2}.$		(7.66)

Note that we choose to write the energy in terms of $\mu$ rather than $T$ because linear density is an easily measurable property whereas the tension is not. One important thing to mention is that the dependence on $A^{2}$ is actually general to all forms of waves, not just sinusoidal. We can calculate the power transmitted through a single point by the wave as

$\displaystyle P=E_{\lambda}f$	$\displaystyle=\frac{1}{2}\lambda f\mu A^{2}\omega^{2}$	(7.67)
	$\displaystyle=\frac{1}{2}v\mu A^{2}\omega^{2}$	(7.68)
	$\displaystyle=\frac{1}{2}\sqrt{\mu T}A^{2}\omega^{2}.$	(7.69)

7.4.2 Power Transmitted through a Wave

We can calculate the power transmitted through the wave from first principles as well. Each string segment exerts a force and does work on the adjoining segments. For a point $x_{0}$ on the string, work is done on the string by to the right of $x_{0}$ by a tension force $T_{y}$ applied by the string to the left of $x_{0}$ . Assuming motion in the $x$ -direction is negligible, we have that the work done by the tension is $\operatorname{\mathrm{d}}\!W=T_{y}\operatorname{\mathrm{d}}\!y$ . Since we are assuming $\frac{\partial y}{\partial x}$ is small, $T_{y}=-T\frac{\partial y}{\partial x}$ . Then we have that the instantaneous power is

$\displaystyle P_{\text{inst}}(x,t)=\frac{\operatorname{\mathrm{d}}\!W}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=T_{y}\frac{\partial y}{\partial t}$	(7.70)
	$\displaystyle=-T\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}$	(7.71)
	$\displaystyle=\frac{T}{V}\left(\frac{\partial y}{\partial t}\right)^{2}=\sqrt{% \mu T}\left(\frac{\partial y}{\partial t}\right)^{2},$	(7.72)

where in the last line we have used the wave equation. For a sinusoidal pattern of motion, we have

P_{\text{inst}}(x,t)=\sqrt{\mu T}A^{2}\omega^{2}\sin^{2}(kx-\omega t).

(7.73)

The average value of $\sin^{2}$ is $\frac{1}{2}$ , so the average power transmitted is

P_{\text{avg}}=\sqrt{\mu T}A^{2}\omega^{2},

(7.74)

which is what we derived before.