OLDHW31 - oldhomewk 31 – PAPAGEORGE MATT – Due 4:00 am...

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Unformatted text preview: oldhomewk 31 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 15, sect 1. part 1 of 2 10 points Two massless springs with spring constants 744 N / m and 5022 N / m are hung from a horizontal support. A block of mass 2 kg is suspended from the pair of springs, as shown. The acceleration of gravity is 9 . 8 m / s . 744 N / m 5022 N / m 2 kg When the block is in equilibrium, each spring is stretched an additional Δ x . Then the block oscillates with an amplitude of 40 m, and it passes through its equilibrium point with a speed of 720 m / s. What is the angular velocity of this system? 1. ω = 4 rad / s 2. ω = 7 rad / s 3. ω = 16 rad / s 4. ω = 18 rad / s correct 5. ω = 8 rad / s 6. ω = 15 rad / s 7. ω = 13 rad / s 8. ω = 10 rad / s 9. ω = 5 rad / s 10. ω = 20 rad / s Explanation: Let m = 2 kg , A = 40 m , v = 720 m / s , k 1 = 744 N / m , and k 2 = 5022 N / m . Basic Concepts: Hooke’s law F =- k x = ma = d 2 x dt 2 d 2 x dt 2 + k m x = 0 , (1) whose integral form has a sine function x ( t ) = A sin( ω t + δ ) , (2) v ( t ) ≡ dx dt v ( t ) = ω A cos( ω t + δ ) , and (3) a ( t ) ≡ dv dt a ( t ) =- ω 2 A sin( ω t + δ ) , where (4) ω = radicalbigg k m . (5) The angular velocity ω is the square root of the coefficient of x in Eq. 1. The frequency of oscillation f versus angu- lar frequency ω is f ≡ ω 2 π . (6) k 1 k 2 m Consider the forces from a spring’s point of view. The oscillating mass exerts the same force, F (at some instant in time) on the springs k 1 and k 2 , F = k 1 x 1 ⇒ x 1 = F k 1 F = k 2 x 2 ⇒ x 2 = F k 2 . oldhomewk 31 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 2 Now consider the effective spring constant, k series , where x = x 1 + x 2 , therefore k series = F x = F x 1 + x 2 = F F k 1 + F k 2 = 1 1 k 1 + 1 k 2 = k 1 k 2 k 1 + k 2 . (7) = (744 N / m) (5022 N / m) (744 N / m) + (5022 N / m) = 648 N / m . Solution: The question presents the springs in series, Eq. 6 and Eq. 7, therefore ω series = radicaltp radicalvertex radicalvertex radicalbt k 1 k 2 m bracketleftBig k 1 + k 2 bracketrightBig (8) = radicaltp radicalvertex radicalvertex radicalbt (744 N / m) (5022 N / m) (2 kg) bracketleftBig (744 N / m) + (5022 N / m) bracketrightBig = radicalBigg (648 N / m) (2 kg) = 18 rad / s , and f = ω series 2 π (9) = (18 rad / s) 2 π = 2 . 86479 cycles / s , and T = 1 f = 1 (2 . 86479 cycles / s) = 0 . 349066 s ....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.

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OLDHW31 - oldhomewk 31 – PAPAGEORGE MATT – Due 4:00 am...

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