Waves

For a stretched string, the speed of a transverse wave is

$$$v=\sqrt{\frac{T}{\mu}}$$$

where $\mu=m/L$.

First find linear mass density:

$$$\mu=\frac{2.50}{20.0}=0.125\ \text{kg m}^{-1}$$$Now wave speed:$$$v=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40\ \text{m s}^{-1}$$$Time taken to travel the length of the string:$$$t=\frac{L}{v}=\frac{20.0}{40}=0.5\ \text{s}$$$

So the disturbance reaches the other end in 0.5 s.

The splash is heard after the stone falls and the sound travels back up.

Time to fall from height $h=300\ \text{m}$:

$$$t_1=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 300}{9.8}}=\sqrt{61.22}\approx 7.82\ \text{s}$$$Time for sound to travel upward:$$$t_2=\frac{300}{340}\approx 0.88\ \text{s}$$$Total time:$$$t=t_1+t_2\approx 7.82+0.88=8.70\ \text{s}$$$

So the splash is heard after about 8.7 s.

For a transverse wave on a string,

$$$v=\sqrt{\frac{T}{\mu}}\Rightarrow T=\mu v^2$$$Linear mass density:$$$\mu=\frac{m}{L}=\frac{2.10}{12.0}=0.175\ \text{kg m}^{-1}$$$

Given $v=343\ \text{m s}^{-1}$,

$$$T=0.175\times (343)^2$$$$$$343^2=117649$$$$$$T=0.175\times 117649\approx 20588\ \text{N}$$$

So the required tension is about **$2.1\times 10^4\ \text{N}$**.

Using

$$$v=\sqrt{\frac{\gamma P}{\rho}}$$$

we explain each part:

- Independent of pressure: For a gas at fixed temperature, density is proportional to pressure, i.e. $\rho \propto P$. So in the ratio $P/\rho$, pressure cancels out. Hence sound speed does not depend on pressure.
- Increases with temperature: For a gas, higher temperature makes molecules move faster and the density for a given pressure decreases. From the gas relation, $v$ increases with $\sqrt{T}$.
- Increases with humidity: Humid air has more water vapour, whose molecular mass is less than that of dry air. This lowers the density $\rho$ of air. Since $v \propto 1/\sqrt{\rho}$, the speed of sound increases.

So the speed of sound in air is independent of pressure, increases with temperature, and increases with humidity.

The converse is not always true. A function must depend on $x$ and $t$ through $x\pm vt$ in order to represent a travelling wave, but not every function of that form is automatically a physically acceptable wave.

Examine each:

**(a) $(x-vt)^2$**
This is of the form $f(x-vt)$, so it can represent a travelling wave.

**(b) $\log[(x+vt)/x_0]$**
This is of the form $f(x+vt)$, so it can also represent a travelling wave, provided the argument of the logarithm is positive.

**(c) $1/(x+vt)$**
This is also of the form $f(x+vt)$, so it can represent a travelling wave, except at points where $x+vt=0$.

Thus, all three are mathematically of travelling-wave form. However, they are not sinusoidal waves; they are just travelling disturbances of the form $f(x\pm vt)$.

Use $\lambda=v/\nu$.

Given frequency:

$$$\nu=1000\ \text{kHz}=10^6\ \text{Hz}$$$**(a) Reflected sound in air** Speed remains that of air:$$$\lambda_{air}=\frac{340}{10^6}=3.4\times 10^{-4}\ \text{m}$$$$$$=340\ \mu\text{m}$$$**(b) Transmitted sound in water**$$$\lambda_{water}=\frac{1486}{10^6}=1.486\times 10^{-3}\ \text{m}$$$$$$=1.486\ \text{mm}$$$

So the wavelengths are **$340\ \mu$m in air and $1.486\ \text{mm}$** in water.

Wavelength is

$$$\lambda=\frac{v}{\nu}$$$Given:$$$v=1.7\ \text{km s}^{-1}=1700\ \text{m s}^{-1}$$$$$$\nu=4.2\ \text{MHz}=4.2\times 10^6\ \text{Hz}$$$So,$$$\lambda=\frac{1700}{4.2\times 10^6}=4.05\times 10^{-4}\ \text{m}$$$$$$=0.405\ \text{mm}$$$

The wavelength is 0.405 mm.

Compare the given wave with the standard form:

$$$y=a\sin(kx\pm \omega t+\phi)$$$Here,$$$y=3.0\sin(36t+0.018x+\pi/4)$$$

So:
- amplitude $a=3.0\ \text{cm}$
- $\omega=36\ \text{rad s}^{-1}$
- $k=0.018\ \text{rad cm}^{-1}$
- initial phase $\phi=\pi/4$

Because the wave is of the form $kx+\omega t$, it travels in the **negative $x$-direction**.

Speed:

$$$v=\frac{\omega}{k}=\frac{36}{0.018}=2000\ \text{cm s}^{-1}$$$$$$=20\ \text{m s}^{-1}$$$Frequency:$$$\nu=\frac{\omega}{2\pi}=\frac{36}{2\pi}\approx 5.73\ \text{Hz}$$$

So it is a travelling wave moving leftward with speed 20 m/s, amplitude 3.0 cm, frequency 5.73 Hz, and initial phase at the origin **$\pi/4$**.

For a fixed position $x$, the displacement varies with time as a sinusoidal curve.

For the given wave:

$$$y(x,t)=3.0\sin(36t+0.018x+\pi/4)$$$

So at each fixed $x$, $y$ versus $t$ is a sine curve with the same amplitude and frequency, but different phase.

At:
- $x=0$,
- $x=2\ \text{cm}$,
- $x=4\ \text{cm}$,

the three graphs are sinusoidal, shifted horizontally relative to one another.

In a travelling wave, the oscillatory motion differs from point to point in phase only; the amplitude and frequency remain the same for all points.

So:
- shape: sinusoidal
- different aspect from point to point: phase