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FinalSol

# FinalSol - PHY 126-07 Final Exam Problem 1(20 points A...

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PHY 126-07 Final Exam Problem 1 (20 points) A small ball of mass 2.00 kg is suspended by two wires, wire 1 and wire 2. If two wires make 90 o and wire 2 makes θ =30 o with respect to the vertical, find the tension T 1 and T 2 . θ = 30 o 90 o T 1 wire 1 T 2 wire 2 P Q mg Solution Let’s denote the length of wire 2 by L. Then the condition that the ball does not rotate about the pivot point of wire 2, Q, is: -L mg sin θ + LT 1 =0 (5 points), namely, T 1 =mg sin θ = (2.00 kg)(9.80 m/s 2 )sin30 o =9.80 N (5 points). The condition that the net force is zero in the horizontal direction gives: T 1 cos θ - T 2 sin θ = 0 (5 points), namely, T 2 =T 1 /tan30 o =17.0 N (5 points). (The condition that the net force is zero in the vertical direction gives: T 1 cos θ + T 2 sin θ = mg can be used too.)

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Problem 2 (10+10 points) An innovative musician is making a new music instrument using cylindrical glasses of height h=1 m with some water in them where the top (bottom) of each glass is open (closed). Right now he is adjusting the height of the water in one of the glasses so that sound of middle C note at frequency 256 Hz creates resonance. Assume that the speed of sound is 344 m/s. (a) What is the maximum height of water for the instrument to resonate with middle C note? (b) What is the next lowest frequency of sound that this glass instrument can resonate with if the water level is the same as in Part (a). open Solution (a) This instrument works as a stopped pipe. Therefore the frequencies of possible n-th harmonics are expressed by f n =nv/(4L n ) where n=1,3,5,…, v is the speed of sound, and L n is the height of air column. Since L n =nv/(4f), for f=256 Hz, L 1 becomes shortest at (344 m/s)/(4 x 256 1/s)=0.336 m when n=1. So the maximum water height for middle C note is 1.000 m - 0.336 m=0.664 m. For a higher harmonics, the height of the air column is larger.
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FinalSol - PHY 126-07 Final Exam Problem 1(20 points A...

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