P17_020 - M 1 and obtain M 1 = M 1 µ 2/µ 1 = 500 g 1 5 00...

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20. (a) The tension in each string is given by τ = Mg/ 2. Thus, the wave speed in string 1 is v 1 = r τ µ 1 = s Mg 2 µ 1 = s (500 g) (9 . 8m / s 2 ) 2(3 . 00 g / m) =28 . 6m / s . (b) And the wave speed in string 2 is v 2 = s Mg 2 µ 2 = s (500 g) (9 . 8m / s 2 ) 2(5 . 00 g / m) =22 . 1m / s . (c) Let v 1 = p M 1 g/ (2 µ 1 )= v 2 = p M 2 g/ (2 µ 2 )and M 1 + M 2 = M. We solve for
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Unformatted text preview: M 1 and obtain M 1 = M 1 + µ 2 /µ 1 = 500 g 1 + 5 . 00 / 3 . 00 = 187 . 5 g ≈ 188 g . (d) And we solve for the second mass: M 2 = M − M 1 = 500 g − 187 . 5 g ≈ 313 g ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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