# 10.2 Speed of sound, frequency, and wavelength  (Page 3/4)

 Page 3 / 4

## Calculating wavelengths: what are the wavelengths of audible sounds?

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in $\text{30.0ºC}$ air. (Assume that the frequency values are accurate to two significant figures.)

Strategy

To find wavelength from frequency, we can use ${v}_{\text{w}}=\mathrm{f\lambda }$ .

Solution

1. Identify knowns. The value for ${v}_{\text{w}}$ , is given by
${v}_{\text{w}}=\left(\text{331}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\sqrt{\frac{T}{\text{273}\phantom{\rule{0.25em}{0ex}}\text{K}}}.$
2. Convert the temperature into kelvin and then enter the temperature into the equation
${v}_{\text{w}}=\left(\text{331}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\sqrt{\frac{\text{303 K}}{\text{273}\phantom{\rule{0.25em}{0ex}}\text{K}}}=\text{348}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{m/s}.$
3. Solve the relationship between speed and wavelength for $\lambda$ :
$\lambda =\frac{{v}_{w}}{f}.$
4. Enter the speed and the minimum frequency to give the maximum wavelength:
${\lambda }_{\text{max}}=\frac{\text{348}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{20 Hz}}=\text{17}\phantom{\rule{0.25em}{0ex}}\text{m}.$
5. Enter the speed and the maximum frequency to give the minimum wavelength:
${\lambda }_{\text{min}}=\frac{\text{348}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{20}\text{,}\text{000 Hz}}=0\text{.}\text{017}\phantom{\rule{0.25em}{0ex}}\text{m}=1\text{.}\text{7 cm}.$

Discussion

Because the product of $f$ multiplied by $\lambda$ equals a constant, the smaller $f$ is, the larger $\lambda$ must be, and vice versa.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If ${v}_{\text{w}}$ changes and $f$ remains the same, then the wavelength $\lambda$ must change. That is, because ${v}_{\text{w}}=\mathrm{f\lambda }$ , the higher the speed of a sound, the greater its wavelength for a given frequency.

## Making connections: take-home investigation—voice as a sound wave

Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table. Gently blow near the edge of the bottom of the sheet and note how the sheet moves. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Explain the effects.

Imagine you observe two fireworks explode. You hear the explosion of one as soon as you see it. However, you see the other firework for several milliseconds before you hear the explosion. Explain why this is so.

Sound and light both travel at definite speeds. The speed of sound is slower than the speed of light. The first firework is probably very close by, so the speed difference is not noticeable. The second firework is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears.

You observe two musical instruments that you cannot identify. One plays high-pitch sounds and the other plays low-pitch sounds. How could you determine which is which without hearing either of them play?

Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

## Section summary

The relationship of the speed of sound ${v}_{w}$ , its frequency $f$ , and its wavelength $\lambda$ is given by

${v}_{\text{w}}=\mathrm{f\lambda },$

which is the same relationship given for all waves.

In air, the speed of sound is related to air temperature $T$ by

${v}_{\text{w}}=\left(\text{331}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\sqrt{\frac{T}{\text{273}\phantom{\rule{0.25em}{0ex}}\text{K}}}.$

${v}_{\text{w}}$ is the same for all frequencies and wavelengths.

## Conceptual questions

How do sound vibrations of atoms differ from thermal motion?

When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.

## Problems&Exercises

When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s?

0.288 m

What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?

Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m.

332 m/s

(a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in [link] is this likely to be?

Show that the speed of sound in $\text{20.0ºC}$ air is 343 m/s, as claimed in the text.

$\begin{array}{lll}{v}_{\text{w}}& =& \left(\text{331 m/s}\right)\sqrt{\frac{T}{\text{273 K}}}=\left(\text{331 m/s}\right)\sqrt{\frac{\text{293 K}}{\text{273 K}}}\\ & =& \text{343 m/s}\end{array}$

Air temperature in the Sahara Desert can reach $\text{56.0ºC}$ (about $\text{134ºF}$ ). What is the speed of sound in air at that temperature?

Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is $\text{20.0ºC}$ .

0.223

A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)

(a) If a submarine’s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.)

(b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.

(a) 7.70 m

(b) This means that sonar is good for spotting and locating large objects, but it isn’t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means.

A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is $\text{24.0ºC}$ and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.

Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See [link] .) (a) Calculate the echo times for temperatures of $\text{5.00ºC}$ and $\text{35.0ºC}$ . (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)

(a) 18.0 ms, 17.1 ms

(b) 5.00%

(c) This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.

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