Pre-Lab Q1: Adolphin is located 150 m above the sea floor. If it squeaks, how mu

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Pre-Lab Q1: Adolphin is located 150 m above the sea floor. If it squeaks, how much time passes before it hears an echo? The speed of sound in sea water is 1530 m/s. b. 0.196 s c. 0.437 s d. 0.875 s Pre-Lab Q2: If the dolphin in the question above moved to a region of colder water, would the time between his squeaking and hearing his echo increase, decrease, or stay the same? a. Increase b. Decrease c. Stay the same Pre-Lab Q3: In general, what could you do experimentally to produce a different value of wavelength but maintain the same frequency? a. Adjust the amplitude of the wave. b. Adjust the length of the tube the sound wave is travelling. c. Change the medium that the wave is travelling in. d. Break the tuning fork. Pre-Lab Q4: What is the fundamental frequency for a 0.30 m long tube closed at one end. Assume the sound wave is travelling at room temperature with a speed of 343 m/s. a. 1.2 Hz b. 143 Hz c. 286 Hz d. 572 Hz Page 4 PHY 1104 Speed of Sound DEPARTMENT OF PHYSICS AND ASTRONOMY APPALACHIAN STATE UNIVERSITY Pre-Lab Q5: What is the average Sound Pressure vs. Time frequency of the waveform for the sound pressure vs time graph to the right? You will be finding the frequency from similar pictures in your lab, and 0.0 describing how you will do this in Question l a. 320 Hz b. 384 Hz c. 440 Hz 0.00 0.0 0.02 d, 493 Hz ime (s 0.03

Explanation / Answer

Q1

Given dolphin is at a position 150 m form the floor of sea

when it squeaks , the time taken for an echo is

here the sound should travel a distnce of 2d that is 2*150 = 300 m to get an echo

   so speed v= d/t ==>t = d/v = 300/1530 s = 0.1960 s

so the answer is option (b)

Q2
   if the dolphin moved to a region of colder water the time between squeaking and the his echo , decreases

option b ----------->> ANS
  

Q3

   in order to producr the different value of wavelength and same frequency

change the medium that the wave is travelling

option C

Q4

given length of the closed tube is L = 0.3 m

and the wavelength (minimum) is Lambda = 4*L

and the relation between frequency and speed and wavelength is V = lambda*f

  

           f = v/Lambda

           f = 343/(4*0.3) Hz

           f = 285.83 Hz
           f = 286 Hz

so the answer is option C

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