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# 1 - homework 21 – FISHER CHRISTIAN – Due 11:00 pm 1...

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Unformatted text preview: homework 21 – FISHER, CHRISTIAN – Due: Apr 13 2008, 11:00 pm 1 Question 1, chap 15, sect 1. part 1 of 1 10 points When a metal ball of unknown mass M is suspended from a spring of unknown force constant k , the spring’s equilibrium length increases by Δ L e . And when the ball is out of equilibrium, it oscillates up-and-down with a period T . Given g = 9 . 8 m / s 2 and T = 0 . 542 s. Find Δ L e . Correct answer: 7 . 29231 cm (tolerance ± 1 %). Explanation: Let L be the length of the free spring, with- out the ball. When a ball is suspended in equi- librium from the spring, the spiring lengthens by Δ L e to generate the tension force F S e = k × Δ L e = Mg, (1) hence Δ L e = Mg k . (2) Now consider the oscillating ball. When the ball is at height y above the equilibrium point, the spring is stretched by Δ L ( y ) = Δ L e- y (3) and therefore has tension F S ( y ) = k × Δ L ( y ) = k × (Δ L e- y ) = Mg- k × y. (4) Consequently, the net vertical force on the ball is F net y = F S- Mg =- k × y, (5) which makes the ball oscillate with angular frequency ω = radicalbigg k M (6) and period T = 2 π ω = 2 π radicalbigg M k . (7) We do not know the ball’s mass M or the spring’s force constant k , but given the oscil- lation period T we may find the ratio M k = parenleftbigg T 2 π parenrightbigg 2 , (8) and therefore Δ L e = Mg k = gT 2 4 π 2 = 7 . 29231 cm . (9) Question 2, chap 15, sect 1. part 1 of 4 10 points An M = 11 . 6 kg mass is suspended on a k = 163000 N / m spring. The mass oscillates up-and-down from the equilibrium position y eq = 0 according to y ( t ) = A sin( ωt + φ ) . Calculate the angular frequency ω of the oscillating mass. Correct answer: 118 . 54 s − 1 (tolerance ± 1 %). Explanation: When the mass moves out of equilibrium, it suffers net restoring force F net y = F spring- Mg =- k ( y- y eq ) =- ky and accelerates back towards the equilibrium position at the rate a y = F net y M =- k M × y. Therefore, the mass oscillates harmonically with angular frequency ω = radicalbigg- a y y = radicalbigg k M = 118 . 54 s − 1 . Question 3, chap 15, sect 1. part 2 of 4 10 points At time t = 0 the mass happens to be at y = 8 . 09 cm and moving upward at velocity v = +36 . 4 m / s. (Mind the units!) homework 21 – FISHER, CHRISTIAN – Due: Apr 13 2008, 11:00 pm 2 Calculate the amplitude A of the oscillating mass. Correct answer: 31 . 7548 cm (tolerance ± 1 %). Explanation: The mass oscillates according to the SHM equation y ( t ) = A sin( ωt + φ ) , hence is velocity is v y ( t ) = dy dt = Aω cos( ωt + φ ) . At time t = 0, we have y ≡ y ( t = 0) = A sin φ and v = v y ( t = 0) = Aω cos φ , or in other words, A sin φ = y , A cos φ = v ω . (1) Consequently, A 2 = ( A sin φ ) 2 + ( A cos φ )0) 2 = ( y ) 2 + parenleftBig v ω parenrightBig 2 , and hence the amplitude A = radicalbigg ( y ) 2 + parenleftBig v ω parenrightBig 2 = 31 . 7548 cm ....
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1 - homework 21 – FISHER CHRISTIAN – Due 11:00 pm 1...

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