phys 101 learning object
TRANSCRIPT
PHYS 101 Learning Object: Sound Wave Speed & Phase of
Medium Damian Feldman-‐Kiss
12036133
Sound Wave Speed & Phase of Medium
Do sound waves travel faster through water or through air? Make a predicLon. Now that you have your predicLon, lets solve a simple physics problem and explore the science behind the phenomenon!
QuesLon Knowing that your physics midterm is coming up, you want to have some fun with science to keep your brain acLve. You decide to see how fast sound travels in the dry air outside the pool versus in the water. You and your pal shout at each other from 50 m apart outside the pool. You then head inside, get changed and try again, but this Lme under water. Calculate the Lme it takes for the sound waves to travel to your pal in each case. Assume temperature is 20˚C for both scenarios. Some helpful values follow. Report your answer with 3 sig. figs. to emphasize the difference in Lmes.
Useful Info
Approximate velocity of a sound wave in dry air: v = 331 ms-‐1 + (0.6 m(s C)-‐1) x T (T measured in Celsius) v = √(B / ρ) See below for soluLon
Bulk modulus (Pa)
Density (kg/m3)
Water(l) 20˚C
2.2 x 109
1 000
SoluLon
Strategy • What are we given and what do we need to solve?
• Visualize problem • Manipulate equaLons and solve
Step 1: Given and unknown values
BWater(l) = 2.2 x 109 Pa ρWater(l) = 1 000 kg/m3
vWater(l) = ? T = 20˚C vAir = ? d = 50 m tAir = ? tWater(l) = ?
Step 2: Draw a Picture
Air:
Water(l):
Hey!!
50 m
50 m
Hey!!
Step 3: EquaLons
Velocity of a sound wave: v = √(B/ρ)
Approximate velocity of a sound wave in dry air: v = 331 ms-‐1 + (0.6 m(s C)-‐1) x T Time: v = d/t => t = d/v
Step 4: Solve Unknown Values
vWater(l) = √(B/ρ) tWater(l) = d/v => d/(√(B/ρ)) = 50m(√(2.2 x 109 Pa / 1 000 kg/m3)) = 0.0337 s vsound wave in dry air = 331 ms-‐1 + (0.6 m(s C)-‐1) x T tAir = d/v => d/(331 ms-‐1 + (0.6 m(s C)-‐1) x 20 C) = 0.146 s
Step 5: Final Answer
At 20˚C, it takes approximately 0.0337 s for sound waves to travel 50 m in liquid water, and approx. 0.146 s in dry air. In other words, sound waves travel approx. 4 Lmes faster in liquid water than in air under these condiLons.
Learning Goal 1
As we just observed, the propagaLon speed of a sound wave depends on the properLes of the medium. Let’s examine this in more detail. (Dr.’s Bates & Ropler, 2015, Lecture 17)
Learning Goal 2 – Why? A sound wave is a longitudinal wave—as the wave propagates through an elasLc medium, the medium alternates between regions of compression and rarefacLon. The wave speed (v) equals the square root of the bulk modulus (B) divided by the density of the medium (p):
v = √(B/ρ)
The bulk modulus is defined as “the raLo of the change in pressure divided by the fracLonal change in volume.” (Hawkes et al., 2014, p. 425)
Learning Goal 2 – Why? The bulk modulus describes how resistant a medium is to compression. Liquids are nearly incompressible, while gases are compressible. Liquids are denser than gases. Sound waves generally propagate faster through a liquid medium than a gaseous medium because of the difference in the compressibility and density between the two phases. But wait, what about the concomitant increase in density from gases to liquids? Would this not compensate for the difference in bulk modulus between the two phases? It turns out that liquids are indeed so resistant to compression that this is not the case. Lets look at another quesLon to drive this point home. (Hawkes et al., 2014, p. 425-‐426)
QuesLon 2
Mercury is a very dense liquid (ρ = 13 534 kg/m3). Its bulk modulus is 2.85 x 1010 Pa. Determine the speed of sound in liquid mercury.
SoluLon
v = √(B/ρ) v = √(2.85 x 1010 Pa / 13 534 kg/m3) v = 1 450 m/s The speed of sound in liquid mercury is 1 450 m/s.
References
Dr.’s Bates and Ropler. 2015. PHYS 101, 202 Lecture Slides. Hawkes et al. 2014. Physics for ScienLsts and Engineers: An interacLve Approach. Revised Custom Volume 1: PHYS 101. Custom Ed UBC.