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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 30.0ºC air. (Assume that the frequency values are accurate to two significant figures.)

Strategy

To find wavelength from frequency, we can use v w = size 12{v size 8{w}=fλ} {} .

Solution

  1. Identify knowns. The value for v w size 12{v size 8{w}} {} , is given by
    v w = 331 m/s T 273 K . size 12{v size 8{w}= left ("331"" m/s" right ) sqrt { { {T} over {"273"" K"} } } } {}
  2. Convert the temperature into kelvin and then enter the temperature into the equation
    v w = 331 m/s 303 K 273 K = 348 . 7 m/s .
  3. Solve the relationship between speed and wavelength for λ size 12{λ} {} :
    λ = v w f . size 12{λ= { {v size 8{w}} over {f} } } {}
  4. Enter the speed and the minimum frequency to give the maximum wavelength:
    λ max = 348 . 7 m/s 20 Hz = 17 m . size 12{λ size 8{"max"}= { {"348" "." 7" m/s"} over {"20 Hz"} } ="17"" m"} {}
  5. Enter the speed and the maximum frequency to give the minimum wavelength:
    λ min = 348 . 7 m/s 20 , 000 Hz = 0 . 017 m = 1 . 7 cm . size 12{λ size 8{"min"}= { {"348" "." 7" m/s"} over {"20" "." "000 Hz"} } =0 "." "0174"" m"=1 "." "74 cm"} {}

Discussion

Because the product of f size 12{f} {} multiplied by λ size 12{λ} {} equals a constant, the smaller f size 12{f} {} is, the larger λ size 12{λ} {} 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 w size 12{v size 8{w}} {} changes and f size 12{f} {} remains the same, then the wavelength λ size 12{λ} {} must change. That is, because v w = , 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 size 12{v size 8{w}} {} , its frequency f size 12{f} {} , and its wavelength λ size 12{λ} {} is given by

v w = ,

which is the same relationship given for all waves.

In air, the speed of sound is related to air temperature T size 12{T} {} by

v w = 331 m/s T 273 K .

v w size 12{v size 8{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 20.0ºC air is 343 m/s, as claimed in the text.

v w = ( 331 m/s ) T 273 K = ( 331 m/s ) 293 K 273 K = 343 m/s alignl { stack { size 12{v rSub { size 8{w} } = \( "331 m/s" \) sqrt { { {"T " \( K \) } over {"273 K"} } } = \( "331 m/s" \) sqrt { { {"293 K"} over {"273 K"} } } } {} #= {underline {"343 m/s"}} {} } } {}

Air temperature in the Sahara Desert can reach 56.0ºC (about 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 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 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 5.00ºC and 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.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
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Abhi
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Source:  OpenStax, Physics of the world around us. OpenStax CNX. May 21, 2015 Download for free at http://legacy.cnx.org/content/col11797/1.1
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