USE THE FOLLOWING CONSTANTS (You may not need to use all of them for this proble

Question

USE THE FOLLOWING CONSTANTS
(You may not need to use all of them for this problem)
• Unless I mention it, these experiments occur on the surface of the earth
and the atmosphere is ’air’.
• Gravity on earth is 10 meters per second squared.
• Gravity on the moon is one sixth that of earth’s gravity.
• Speed of sound in air is 350 meters per second.
• Density of water is 1000 kilograms per cubic meter.
• Density of mercury is 15000 kilograms per cubic meter.
• Density of ethyl alcohol is 750 kilograms per cubic meter.
All answers must come with explanations. Merely guessing will not give you
full credit.

5 Floating Tube
You have a hollow cylinder open at both ends made from the element ’Rebel-
sium’. Its inner radius is 3 centimeters and its outer radius is 4 centimeters
and is 2.3 meters long. When placed in water, it floats perfectly upright but
sinks partially into the water. It can now produce a fundamental frequency
of 300 Hz. What is the density of ’Rebelsium’? What is the frequency of the
tubes first overtone if the tube is now resting upright in mercury instead of
water?

Explanation / Answer

Given: inner radius of the tube r = 0.03 m outer radius of the tube R = 0.04 m heitht of the tube h = 2.3 m fundamental frequency produced by tube = 300 Hz. fundamental frequency of open pipe open at both ends f = v/4L ...... (1) here L is the length of the tube L = 2.3 m now from the equation 1 v = f*4L    = (300 Hz)(4* 2.3 m )    = 2760 m/s speed of the sound wave in that tube = 2760 m/s now from the speed of sound in a liquid or gas having a bulk modulus B and density is gievn by v = sqrt(B/) .......   (2) from the above relation density of Rebelsium = v2/B ...... (3) plug the Bulk modulus of given element in equation 3 to get density. (ii) from the equation 2 v = sqrt (B/) here B is the Bulk modulus of Mercury = 28.5*109 Pa. N/m2 is the density of Mercury = 15*103 kg/m3 v = sqrt (28.5*109 Pa. N/m2 / 15*103 kg/m3)     = 1.3*103 m/s frequecy of the first overtone f = v/4L                                                   = (1378 m/s)/4*2.3                                                   = 149.8 Hz                                                   = 150 Hz nearly.   

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