Quiz_4_Solutions

Quiz_4_Solutions - Ts of 121 N and 129 N. ____ _____ 5....

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Quiz 4 Solutions 1. For a closed pipe successive harmonics differ by 2f 1 as only odd harmonics appear. So 2f 1 = 630 – 450 = 180 giving f1 = 90 Hz. 2. For a closed pipe, L = nλ/4 (n odd) so λ = 4L/n = v/f n . This gives for n = 1 f 1 = v/4L and substituting gives L = 343/4 x 90 = 0.95 m. 3. For a string fixed at both ends, L = nλ/2 which, as above, becomes f 1 = v/2L where ___ for a string v = √T/μ where μ is the linear density of the string, m/L. Here μ= 0.65x10 -3 kg/0.7 m = 9.29 x 10 -4 kg/m. Entering the formula for v and solving for T gives T = (2Lf 1 ) 2 μ. Putting in the numbers gives T 125 N. 4. Since │f 1 – f 2 │= f beat the two possible frequencies for the other player are f = 262 ± 4 or 258 and 266 as the possible frequencies. Putting these into the tension formula from 3 gives
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Unformatted text preview: Ts of 121 N and 129 N. ____ _____ 5. Since f = (1/2)K/m, entering the numbers gives f = (1/2)200/5 = 1.01 Hz. 6. There are two approaches here. ____ a) v max = A where A = 20 cm or 0.2 m and = K/m . The numbers give v max = 1.26 m/s. b) 1/2mv 2 = 1/2KA 2 where the velocity at x = 0 is the maximum velocity. Here ______ ____ v = KA 2 /m = AK/m = 1.26 m/s. 7. Here, no matter the direction of the displacement, one spring pulls and the other pushes so their individual forces are in the same direction, back to the equilibrium position. So the forces add producing a net force whose magnitude is 2Kx, like a spring with double the K. _____ _____ Therefore f = (1/2)2K/m = (1/2)400/5 = 1.4 Hz....
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