oldmidterm4 #1 - hinojosa(jlh3938 – oldmidterm4 01 –...

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Unformatted text preview: hinojosa (jlh3938) – oldmidterm4 01 – Turner – (58185) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A mass attached to a spring executes sim- ple harmonic motion in a horizontal plane with an amplitude of 4 . 97 m. At a point 3 . 1311 m away from the equilibrium, the mass has speed 4 . 65 m / s. What is the period of oscillation of the mass? Consider equations for x ( t ) and v ( t ) and use sin 2 + cos 2 = 1 to calculate ω . Correct answer: 5 . 21528 s. Explanation: Let A = 4 . 97 m , x = 3 . 1311 m , and v = 4 . 65 m / s . The simplest solution uses the equation for simple harmonic motion x ( t ) = A sin( ω t ) , and its time derivative v ( t ) = dx dt = Aω cos( ω t ) . For any angle (such as the phase angle ω t ) sin 2 ( ω t ) + cos 2 ( ω t ) = 1 , so parenleftBig x A parenrightBig 2 + parenleftBig v A ω parenrightBig 2 = 1 , and parenleftBig v ω parenrightBig 2 = A 2 − x 2 . Consequently, ω = v √ A 2 − x 2 = 4 . 65 m / s radicalbig (4 . 97 m) 2 − (3 . 1311 m) 2 = 1 . 20476 s − 1 and T = 2 π ω = 2 π 1 . 20476 s − 1 = 5 . 21528 s . 002 10.0 points A block of unknown mass is attached to a spring of spring constant 7 . 1 N / m and under- goes simple harmonic motion with an ampli- tude of 8 . 5 cm. When the mass is halfway between its equilibrium position and the end- point, its speed is measured to be 21 . 4 cm / s. Calculate the mass of the block. Correct answer: 0 . 840098 kg. Explanation: Let : k = 7 . 1 N / m , A = 8 . 5 cm , and v = 21 . 4 cm / s . If the maximum displacement (amplitude) is A , the halfway displacement is A 2 . By energy conservation, K i + U i = F f + U f 0 + 1 2 k A 2 = 1 2 mv 2 + 1 2 k parenleftbigg A 2 parenrightbigg 2 k A 2 = mv 2 + 1 4 k A 2 m = 3 k A 2 4 v 2 = 3 (7 . 1 N / m) (0 . 085 m) 2 4 (0 . 214 m / s) 2 = . 840098 kg . 003 10.0 points A hoop ( i.e. , a ring) of radius 3 m and mass 13 kg is suspended from a pivot on the perime- ter of the hoop as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . x y θ hinojosa (jlh3938) – oldmidterm4 01 – Turner – (58185) 2 Find the angular frequency ω of small os- cillations. Correct answer: 1 . 27802 rad / s. Explanation: Basic Concepts T = 2 π radicalBigg I mg d Solution: From the parallel axis theorem, the moment of inertia of the hoop about the pivot is given by I = I c + M d 2 = mR 2 + mR 2 = 2 mR 2 Hence the angular frequency is ω = radicalbigg mg d I = radicalbigg mg R 2 mR 2 = radicalbigg g 2 R = 1 . 27802 rad / s . 004 10.0 points A 0 . 309 kg mass is attached to a spring and undergoes simple harmonic motion with a pe- riod of 0 . 328 s. The total energy of the system is 4 . 6 J....
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oldmidterm4 #1 - hinojosa(jlh3938 – oldmidterm4 01 –...

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