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: oldmidterm 04 PAPAGEORGE, MATT Due: Apr 29 2008, 4:00 am 1 Question 1, chap 15, sect 1. part 1 of 1 10 points Hint: Write down equations for x ( t ) and v ( t ) and use sin 2 +cos 2 = 1 to calculate . A mass attached to a spring executes sim ple harmonic motion in a horizontal plane with an amplitude of 0 . 337 m. At a point . 27297 m away from the equilibrium, the mass has speed 3 . 32 m / s. What is the period of oscillation of the mass? Correct answer: 0 . 374014 s (tolerance 1 %). Explanation: There are several ways to solve this prob lem for example, one may use the energy conservation but the simplest solution uses nothing but the equation of simple harmonic motion x ( t ) = A sin( t ) , and its time derivative v ( t ) = dx dt = A cos( t ) . For any angle [such as the phase angle t ] sin 2 ( t ) + cos 2 ( t ) = 1 , hence parenleftBig x A parenrightBig 2 + parenleftBig v A parenrightBig 2 = 1 , and therefore parenleftBig v parenrightBig 2 = A 2 x 2 . Consequently, = v A 2 x 2 = 16 . 7993 s 1 . Finally, the period of the oscillation follows from the angular frequently, T = 2 = . 374014 s . Question 2, chap 15, sect 3. part 1 of 1 10 points A block of unknown mass is attached to a spring of spring constant 9 . 92 N / m and undergoes simple harmonic motion with an amplitude of 9 . 02 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 39 . 2 cm / s. Calculate the mass of the block. Correct answer: 0 . 393925 kg (tolerance 1 %). Explanation: Basic Concepts Energy conservation: If K is kinetic energy and U is potential energy, K i + U i = K f + U f Kinetic energy of particle with mass m and speed v : K = 1 2 mv 2 Mass m on spring with constant k : = radicalbigg k m and potential energy of a spring at displace ment x : U = 1 2 k x 2 Period T = 2 Solution: Call the maximum displacement (amplitude) A . The halfway displacement is A/ 2. Energy conservation requires 0 + 1 2 k A 2 = 1 2 mv 2 + 1 2 k parenleftbigg A 2 parenrightbigg 2 or k A 2 = mv 2 + 1 4 k A 2 so m = 3 k A 2 4 v 2 = (3) (9 . 92 N / m) (0 . 0902 m) 2 (4) (0 . 392 m / s) 2 = . 393925 kg . oldmidterm 04 PAPAGEORGE, MATT Due: Apr 29 2008, 4:00 am 2 Question 3, chap 11, sect 3. part 1 of 1 10 points A thin hoop of radius 2 . 56 m and mass 9 . 5 kg is suspended from a pivot on the hoop itself as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . Find the period T of small oscillations of this hoop around = 0. Correct answer: 4 . 54153 s (tolerance 1 %). Explanation: The hoop oscillates as a physical pendulum. Its center of mass is in its geometric center, at distance R = 2 . 56 m from the pivot. So when it swings by a small angle away from the equilibrium position (center directly be low the pivot), there is a torque due to gravity force, = mg R sin mg r (1) for small ....
View
Full
Document
 Spring '09
 KLEINMAN
 Physics, Mass

Click to edit the document details