Chapter 10: Sound Waves: Characteristics and Applications (Physics)

Answer: (ii)

A mechanical wave travels by the vibration of particles of a material medium, so it cannot travel through a vacuum. Sound needs a medium (solid, liquid or gas), which is what makes it mechanical. Reflection, frequency and carrying energy are properties of all waves, including light, which is not a mechanical wave, so they do not show this.

(ii) Sound needs a medium to propagate.

Answer: (iii)

The speed of sound depends on the medium, not on the frequency, so $v=f\lambda$ stays the same. A higher frequency therefore means a shorter wavelength and a shorter time period ($T=\frac{1}{f}$). Frequency is the number of compressions (or rarefactions) passing a point per second, so this increases.

(iii) number of compressions per second

$$f=\frac{\text{number of compressions}}{\text{time}}=\frac{20}{4}=5\text{Hz}$$

(ii) 5 Hz

It will produce reverberation, not an echo. The sensation of a sound stays in our brain for about 0.1 s. To be heard as a separate echo, the reflected sound must arrive at least 0.1 s after the original. Here it arrives after only 0.05 s, so it merges with the original sound and makes it seem prolonged; repeated reflections from the walls make the sound persist. This is reverberation.

Reverberation, because 0.05 s is less than 0.1 s, so the reflected sound blends with the original sound instead of being heard separately.

(i) Wave (a). Over the same distance, (a) shows fewer (about 3) waves while (b) shows about 6, so each wave of (a) is longer: greater wavelength.

(ii) Wave (a). Its crests rise less above the mean (dashed) line than those of (b), so its amplitude is smaller.

(i) (a) has the greater wavelength (ii) (a) has the smaller amplitude

All three waves travel in the same medium, so they have the same speed. Since $v=f\lambda$, the highest frequency means the shortest wavelength (the most waves in the same distance).

• A = the curve with the most waves over the distance (shortest wavelength); the green curve in Fig. 10.31.
• B = the curve in between; the red curve.
• C = the curve with the fewest waves (longest wavelength); the blue curve.

A: the curve with the most (closest) waves (green); B: the middle one (red); C: the one with the fewest, longest waves (blue).

Take the normal density of the medium as the zero line. The graph is a wave that rises to +3 units (compressions) and falls to −3 units (rarefactions), repeating every 4 cm (crests at 1 cm, 5 cm, 9 cm…).

A wave of height 3 units above and below the normal-density line, with crests 4 cm apart (see graph).

• Sound cannot be heard in space. Space is a near vacuum, with no medium for sound to travel through (sound is a mechanical wave). Only the flash of light (which can travel through vacuum) would be seen; the explosion would be silent.
• Light and sound would not arrive together anyway. Even where there is a medium (like air), light travels about $3\times 10^8$ m s⁻¹ while sound travels only about 340 m s⁻¹. So the flash would be seen first and the sound heard later, as with lightning and thunder.

There would be no sound in space (no medium), and even in air the flash would be seen before the sound is heard, since light is much faster than sound.

$$f=\frac{v}{\lambda }=\frac{344}{3.44}=100\text{Hz},T=\frac{1}{f}=\frac{1}{100}=0.01\text{s}$$

(Or directly, $T=\frac{\lambda }{v}=\frac{3.44}{344}=0.01$ s.)

0.01 s

The signal travels down to the wreck and back up, so it covers twice the depth $d$:

$$2d=v\times t=1525\times 5=7625\text{m}\Rightarrow d=3812.5\text{m}$$

About 3812.5 m (≈ 3.8 km) below the ship.

Total distance $=2\times 1.2=2.4$ m.

$$t=\frac{2.4}{345}\approx 0.007\text{s}(\approx 7\text{ms})$$

About 0.007 s (7 milliseconds)

$$t_{22°\text{C}}=\frac{1720}{344}=5\text{s},t_{0°\text{C}}=\frac{1720}{331}\approx 5.196\text{s}$$

$$\text{Extra time}\approx 5.196-5=0.196\text{s}\approx 0.2\text{s}$$

(Exactly, $\frac{1720}{331}-5=\frac{65}{331}$ s.) Sound travels slower in colder air.

About 0.2 s (65/331 s) more.

In Fig. 10.32 the 8 cm marked distance runs from the centre of one compression (dense region) to the centre of the third compression, so it covers two wavelengths:

$$\lambda =\frac{8}{2}=4\text{cm}=0.04\text{m}$$

$$f=\frac{v}{\lambda }=\frac{340}{0.04}=8500\text{Hz}$$

Wavelength = 0.04 m (4 cm); frequency = 8500 Hz

From Fig. 10.33, wave A completes one full wave every 2.5 cm and wave B every 5.0 cm.

$${\lambda }_A=2.5\text{cm}=0.025\text{m},f_A=\frac{345}{0.025}=13800\text{Hz}$$

$${\lambda }_B=5.0\text{cm}=0.05\text{m},f_B=\frac{345}{0.05}=6900\text{Hz}$$

(B has twice the wavelength, so half the frequency of A.)

A: 0.025 m, 13,800 Hz; B: 0.05 m, 6900 Hz

Both sounds travel the same distance $2d$ (to the cliff and back). With $t_A=4.5t_B$:

$$v_{\text{air}}=\frac{2d}{t_A},v_{\text{water}}=\frac{2d}{t_B}\Rightarrow \frac{v_{\text{air}}}{v_{\text{water}}}=\frac{t_B}{t_A}=\frac{1}{4.5}=\frac{2}{9}$$

$v_{\text{air}}:v_{\text{water}}=2:9$