Review Practice Questions Answer Key 171x5b

Review Practice Questions Answer Key: 1. A person sitting in a boat attempting to catch fish counts 18 waves ing by the boat every minute. If the distance between crests is 1.4 m determine the speed of the wave. 𝑐𝑦𝑐𝑙𝑒𝑠 18 𝑓𝑟𝑒𝑞 = = = 0.30 𝐻𝑧 𝑡𝑖𝑚𝑒 60 𝑠𝑒𝑐 𝑣 = 𝑓𝜆 = 0.30 𝐻𝑧 1.4𝑚 𝒗 = 𝟎. 𝟒𝟐 𝒎/𝒔 2. 3.

Give an example of a longitudinal wave , a transverse wave, and an object that can function as either one. Examples of longitudinal waves: sound waves, tsunami waves, earthquake P waves, ultra sounds, vibrations in gas, and oscillations in spring, internal water waves, and waves in slinky Examples of Transverse waves: Radio waves, light waves, thermal radiation, X ray, slinky, rope, water waves. Both: Slinky, water waves.

Sketch the pulses 4 seconds later:





4.

Sketch the 3rd overtone of a closed end tube in the diagram below

5.

Laser sighting can be used to calculate the apparent distance to an object. A laser in air (at 25° from the normal) detects an impurity in a sample of glass and records the apparent depth as 2.34 m. Calculate the actual depth of the object.



𝑔𝑙𝑎𝑠𝑠 ℎ𝑎𝑠 𝑎 𝑜𝑝𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 1.50 𝑛! 𝑆𝑖𝑛(𝜃! ) = 𝑛! 𝑆𝑖𝑛(𝜃! )



𝑆𝑖𝑛(𝜃! ) =



𝑆𝑖𝑛(𝜃! ) =



𝑆𝑖𝑛(25)1.00 1.50

𝜃! = 16.4° 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ 𝑇𝑎𝑛(25°) =

𝑇𝑎𝑛(16.4°) =

𝑜𝑝𝑝 𝑜𝑝𝑝 = 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡ℎ 2.34 𝑚 𝑂𝑝𝑝 = 1.09 m



𝑆𝑖𝑛(𝜃! )𝑛! 𝑛!

𝑜𝑝𝑝 1.09 = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑑𝑒𝑝𝑡ℎ 𝐷

𝑫 = 𝟑. 𝟕𝟏 𝒎

6.

Two friends are practicing dribbling basketballs at the same rate but completely out of phase from each other. a) A third friend standing a distance away hears nothing, Is this friend on a nodal line or antinodal line? Nodal b) If the third friend determines that they are one the 2nd _____________ line and is from part “a” 7m away from the one friend and 8.2m away from the second friend, determine the wavelength of the sound.

a) 𝑅𝑒𝑔𝑎𝑟𝑑𝑙𝑒𝑠𝑠 𝑜𝑓 𝑤ℎ𝑒𝑡ℎ𝑒𝑟 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑖𝑛 𝑝ℎ𝑎𝑠𝑒 𝑜𝑟 𝑜𝑢𝑡 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒, 𝑛𝑜𝑑𝑎𝑙 𝑙𝑖𝑛𝑒𝑠 𝑐𝑜𝑟𝑟𝑖𝑠𝑝𝑜𝑛𝑑 𝑡𝑜 𝑎𝑟𝑒𝑎𝑠 𝑜𝑓 𝑑𝑒𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑒𝑛𝑐𝑒, 𝑡ℎ𝑢𝑠 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑓𝑟𝑖𝑒𝑛𝑑 𝑖𝑠 𝑜𝑛 𝑎 𝑛𝑜𝑑𝑎𝑙 𝑙𝑖𝑛𝑒. b) 𝑜𝑢𝑡 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒 𝑛𝑜𝑑𝑎𝑙 𝑙𝑖𝑛𝑒 𝑃𝐷: 𝑃𝐷 = 𝑛𝜆 8.2 𝑚 − 7 𝑚 = 2𝜆 1.2 𝑚 = 2𝜆 𝟎. 𝟔𝟎 𝒎 = 𝝀 7. Given that the light moves at a speed of 3.00 x 108 m/s, determine the n value for the unknown medium and find the velocity of light in that medium. 𝑆𝑛𝑒𝑙𝑙 ! 𝑠 𝐿𝑎𝑤: 𝑛! 𝑠𝑖𝑛𝜃! 𝑣! = = 𝑛! 𝑠𝑖𝑛𝜃! 𝑣! 40.9˚ 1.00 Sin 32.0° = 𝑛! 𝑆𝑖𝑛 40.9°

air

𝒏𝟐 = 𝟏. 𝟐𝟒 𝑠𝑖𝑛𝜃! 𝑣! = 𝑠𝑖𝑛𝜃! 𝑣! Sin 32.0° 𝑣! = 𝑆𝑖𝑛 40.9° 3.00×10! 𝑚/𝑠 𝒗𝟐 = 𝟐. 𝟒𝟑×𝟏𝟎𝟖 𝒎/𝒔

32.0˚

8.

Which wave will diffract more? Explain your reasoning.

Width of aperture and wavelength contribute to the diffraction. We want a small aperture and a large wavelength. This means diagram 1 will diffract more. 9.

A student finds an unlabeled tuning fork and attempts to determine the frequency of the fork. They set up the lab equipment used during the resonance lab and find lengths of tube (closed tube) that produce loud noises. The first loud noise is detected at a length of 0.184 m a) What is the frequency of the tuning fork if the temperature in the room is exactly 20.0o C b) The student measured the next loud point at a length of 0.552 m. Does this agree with the data collected and our understanding of closed tubes standing waves? Bonus: What note is this tuning fork tuned to? a) 𝑚 𝑣! = 343 𝑠 4𝑙 4 0.184 𝑚 𝜆= = = 0.736 𝑚 𝑛 1 𝑓= b)

𝑣 343 = = 𝟒𝟔𝟔 𝑯𝒛 𝜆 0.736 𝑚

𝑁𝑒𝑥𝑡 𝑙𝑜𝑢𝑑 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 3𝑟𝑑 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐: 𝑛 = 3 𝑛𝜆 3 0.736 𝑚 𝑙= = = 0.552 𝑚 4 4 𝒀𝒆𝒔 𝒕𝒉𝒊𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒆𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒍𝒆𝒏𝒈𝒕𝒉 𝒕𝒐 𝒂𝒄𝒉𝒊𝒆𝒗𝒆 𝒕𝒉𝒆 𝒕𝒉𝒊𝒓𝒅 𝒉𝒂𝒓𝒎𝒐𝒏𝒊𝒄. Bonus: B flat

An ambulance is driving toward a person on the sidewalk. Typically sirens emit a frequency of 1.300 x 103 Hz. The observer hears a frequency of 1.393 x 103 Hz. What speed is the ambulance travelling at?

𝑓! = 𝑓

𝑣 𝑣 − 𝑣!

𝑓! 𝑣 = 𝑓 𝑣 − 𝑣! 𝑓𝑣 = 𝑣 − 𝑣! 𝑓! 𝑣! = 𝑣 −

𝑓𝑣 𝑓!

𝑣! = 𝑣 1 − 𝑣! = 343 𝑚/𝑠 1 −

10.

11.

𝑓 𝑓!

1.300×10! 𝐻𝑧 1.393×10! 𝐻𝑧

𝒗𝒔 = 22.90 m/s Two tuning forks with slightly different frequencies are played at the same time and a beat frequency of 6 is heard. The lower frequency tuning fork in inscribed with the number 304. What is the frequency of the second turning fork? 𝑇ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑏𝑒𝑎𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑥 − 304 = 6 𝒙 = 𝟑𝟏𝟎 𝑯𝒛 Given the pressure vs. distance graph for a longitudinal wave, determine the wavelength of the wave.

λ

77𝑐𝑚 − 42 𝑐𝑚 = 𝜆 35 𝑐𝑚 = 𝜆 𝟎. 𝟑𝟓 𝒎 = 𝝀