Figure 17-32 shows the output from a pressure monitor...

Figure $17-32$ shows the output from a pressure monitor mounted at a point along the path taken by a sound wave of a single frequency traveling at 343 $\mathrm{m} / \mathrm{s}$ through air with a uniform density of $1.21 \mathrm{~kg} / \mathrm{m}^{3} .$ The vertical axis scale is set by $\Delta p_{s}=4.0 \mathrm{mPa}$. If the displacement function of the wave is $s(x, t)=s_{m} \cos (k x-\omega t)$, what are (a) $s_{m}$, (b) $k$, and (c) $\omega ?$ The air is then cooled so that its density is $1.35 \mathrm{~kg} / \mathrm{m}^{3}$ and the speed of a sound wave through it is $320 \mathrm{~m} / \mathrm{s}$. The sound source again emits the sound wave at the same frequency and same pressure amplitude. What now are (d) $s_{m}$ (e) $k$, and (f) $\omega$ ?

Key Concepts

Sound Wave Propagation

Sound waves are mechanical disturbances that travel through a medium by causing oscillatory motion in its particles. Their behavior is modeled by sinusoidal functions which describe how pressure and displacement vary both in space and time. The speed of sound, which is determined by the medium’s properties such as density and elasticity, plays a central role in linking the temporal and spatial characteristics of the wave.

Displacement and Pressure Amplitudes Relationship

In sound waves, there is a direct relationship between the displacement amplitude (the maximum distance particles are displaced) and the pressure amplitude (the maximum pressure variation) via the medium’s density and the wave’s speed and frequency. This relationship is often expressed as ?p = ? v ? s_m, which indicates that for a given pressure amplitude, changes in the medium’s density or sound speed, or in the frequency, will result in different displacement amplitudes.

Wave Number

The wave number, defined as k = 2?/?, where ? is the wavelength, quantifies the spatial rate of change of the phase of a wave. It captures how rapidly the oscillatory behavior of the wave repeats in space and is an essential parameter in characterizing the propagation and interference of waves in various media.

Angular Frequency

Angular frequency, denoted by ? and given by ? = 2? f (with f being the frequency), measures how fast a wave oscillates in time. It is fundamental in the mathematical description of periodic motions, directly linking the time-dependent properties of the wave to its sinusoidal representation.

Effects of Medium Properties on Sound Waves

The properties of the medium, such as density and speed of sound, significantly influence the characteristics of sound waves. While the pressure amplitude might remain constant when a source emits sound at a fixed intensity, changes in the medium alter the displacement amplitude and wave number through their effect on the relationship between pressure variation, particle motion, and wave propagation speed.

Transcript

00:01 For this problem on the topic of waves, we are shown in the figure the output from a pressure monitor that is observing a sound wave of a single frequency traveling at 343 meters per second through a.

00:12 A has a uniform density of 1 .21 kilograms per cubic meter, and the vertical axis scale of the graph is set by delta ps is 4 milipascals.

00:24 Now, if the displacement function of the wave as a function of x and t is, the displacement.

00:30 Amplacement amplitude sm times the cosine of kx minus omega t we want to find the displacement amplitude sm the values for k and the value for omega if it is then cool so that its density changes to 1 .35 kilograms per cubic meter and the speed of a sound wave is now 320 meters per second we want to find the new values for sm k and omega now the period t we can see is two milliseconds, which we can write as 0 .002 seconds, and the pressure amplitude delta p .m.

01:17 Is equal to 8 milliseconds.

01:22 This is equivalent to 0 .008 pascels or neutrons per square meter and the displacement amplitude we know sm is equal to the pressure amplitude delta pm over v times row times omega which we can write as delta pm divided by v times row multiplied by 2 pi over t and so substituting the values above we get this displacement amplitude to be 6 .1 times 10 to the minus 9 meters.

02:19 Next for part b we want to find the angular wave number.

02:24 So the angular wave number k is equal to omega over v, which is 2 pi over v times t.

02:36 We are given the speed of the wave to initially be 343 meters per second.

02:40 So we get the angular wave number to be 9 .2 radians per meter...

Please give Ace some feedback