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Test #1 Sample Problems in CAS Physics

Part I: Matching Terms with Definitions

  • 1) The highest points of a transverse wave.

    • Term: D) Crests

  • 2) How many seconds it takes one energy wave to pass a given point in space.

    • Term: H) Period

  • 3) Sound waves are

    • Term: M) Longitudinal

  • 4) Any substance that has both mass and volume.

    • Term: R) Matter

  • 5) Water waves cause water molecules to vibrate

    • Term: A) Wavelength

  • 6) The pitch humans perceive is based on this property of a sound wave.

    • Term: B) Frequency

  • 7) Type of energy waves that can travel through a vacuum.

    • Term: L) Transverse

  • 8) At colder temperatures, the density of a medium

    • Term: U) Increases

  • 9) How strongly a medium resists compression.

    • Term: J) Bulk Modulus

  • 10) Sound waves with frequencies above 20,000 Hz.

    • Term: J) Ultrasound

Part II: True or False Statements

  • 1) Sound waves cannot travel through a vacuum.

    • Answer: T

  • 2) Constructive interference decreases the amplitudes of sound waves.

    • Answer: F (Corrected to: Constructive interference increases the amplitudes of sound waves.)

  • 3) The Law of Entropy states that energy becomes more useful every time energy changes form.

    • Answer: F (Corrected to: The Law of Entropy states that energy becomes less useful every time energy changes form.)

  • 4) The larger the wavelength, the higher the frequency of a wave.

    • Answer: F (Corrected to: The larger the wavelength, the lower the frequency of a wave.)

  • 5) Sound waves travel fastest through mediums in the gaseous phase.

    • Answer: F (Corrected to: Sound waves travel fastest through solids.)

Part III: Problem Solving Questions

Question 1: Sound from an Explosion

  • Scenario: A swimmer experiences sound from an explosion underwater and in air.

  • Given Data:

    • Time under water: t_w = 1.025 ext{ s}

    • Time in air after exiting water: t_A = 3.425 ext{ s}

    • Speed of sound in saltwater is 1,050 ext{ m/s} faster than in air.

a) Time for sound in air to reach swimmer:

  • Calculation:

    • tA = tw + 3.425

    • t_A = 1.025 + 3.425 = 4.450 ext{ s}

b) Speed of sound in air and saltwater:

  • Establishing equations:

    • d = V_A (4.450)

    • d = (V_A + 1,050) (1.025)

  • Equating distances:

    • 4.450VA = 1.025VA + 1,080

    • 3.425V_A = 1,080

    • VA = rac{1,080}{3.425} ightarrow VA = 315.3 ext{ m/s}

    • VW = VA + 1,050 = 1,365 ext{ m/s}

c) Distance between swimmer and explosion:

  • Using first calculated speed:

  • Underwater Calculation:

    • d = V_A (4.450)

    • d = 315.3 imes 4.450

    • d = 1,403 ext{ m}

Question 2: Piano String and Harmonics

  • Given Data:

    • Linear mass density, = 0.0122 ext{ kg/m}

    • Tension, F_T = 793 ext{ N}

a) Wave speed of standing wave:

  • Formula:

    • V = rac{F_T}{}

  • Calculation:

    • V = rac{793}{0.0122}
      ightarrow V = 255 ext{ m/s}

b) Wavelength of the third harmonic:

  • Given frequency:

    • Frequency of third harmonic, f = 370.5 ext{ Hz}

  • Formula:

    • V = ext{wavelength} imes f

  • Calculation:

    • ext{wavelength} = rac{255}{370.5}
      ightarrow ext{wavelength} = 0.688 ext{ m}

c) Harmonics Chart:

  • Harmonic Number (n)

    • 1: Wavelength = 2.064, Frequency = 123.5 Hz

    • 2: Wavelength = 1.032, Frequency = 247.0 Hz

    • 3: Wavelength = 0.688, Frequency = 370.5 Hz

    • 4: Wavelength = 0.513, Frequency = 494.0 Hz

    • 5: Wavelength = 0.426, Frequency = 617.5 Hz

d) Length of the piano string:

  • Using first harmonic relationship:

  • Formula:

    • L = rac{ ext{wavelength}}{2}

  • Calculation:

    • L = rac{2.064}{2} = 1.032 ext{ m}

Question 3: Sound Waves from Amplifier

  • Given Data: Power emitted by speaker, P = 225 ext{ W}

a) Sound intensity at distance of 1.25 m:

  • Intensity Formula:

    • I = rac{P}{A}

  • Surface Area of Sphere:

    • A = 4 imes ext{π} imes r^2

  • Calculation:

    • A = 4 imes ext{π} imes (1.25^2)

    • I
      ightarrow rac{225}{(4 imes ext{π} imes 1.25^2)}
      ightarrow I = 11.5 = 1.15 imes 10^{-1} ext{ W/m}^2

b) Decibel level at 1.25 m:

  • Decibel Formula:

    • dB = 10 imes ext{log} rac{I}{I_0}

    • Where I_0 = 1 imes 10^{-12} ext{ W/m}^2

  • Calculation:

    • dB = 10 imes ext{log} (1.15 imes 10^{-1} / 1 imes 10^{-12})

    • dB = 131 ext{ dB}

    • Conclusion: 131 dB is damaging to humans.

c) Sound intensity at distance of 3.00 m:

  • Calculation:

    • A = 4 imes ext{π} imes (3.00^2)

    • I = rac{225}{4 imes ext{π} imes (3.00^2)}
      ightarrow I
      ightarrow 1.99 imes 10^{-1} ext{ W/m}^2

d) Decibel level at 3.00 m:

  • Calculation:

    • dB = 10 imes ext{log} (1.99 imes 10^{-1} / 1 imes 10^{-12})

    • dB = 123 ext{ dB}

    • Conclusion: 123 dB is still dangerous!

e) Loudness comparison:

  • Formula to find how many times louder:

    • ext{Times louder} = 10^{( ext{Bel difference})}

    • Bel difference: 131 - 123 = 8

  • Conclusion: Sound at 1.25 m is approximately 6.31 times louder than at 3.00 m.