This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hinojosa (jlh3938) – oldmidterm4 01 – Turner – (58185) 1 This printout should have 21 questions. Multiplechoice 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....
View
Full
Document
This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Mass

Click to edit the document details