HW11 Solution - salazar (ggs332) HW11 TSOI (58160) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: salazar (ggs332) HW11 TSOI (58160) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A body oscillates with simple harmonic mo- tion along the x-axis. Its displacement varies with time according to the equation x ( t ) = A sin( t + ) . If A = 3 m, = 3 . 572 rad / s, and = 1 . 0472 rad, what is the acceleration of the body at t = 4 s? Note: The argument of the sine function is in radians rather than degrees. Correct answer: 13 . 9404 m / s 2 . Explanation: Let : A = 3 m , = 3 . 572 rad / s , = 1 . 0472 rad , and t = 4 s . x = A sin( t + ) v = dx dt = A cos( t + ) a = dv dt = 2 A sin( t + ) = 2 A sin( t + ) = (3 . 572 rad / s) 2 (3 m) sin[(3 . 572 rad / s)(4 s) + 1 . 0472 rad] = 13 . 9404 m / s 2 . 002 10.0 points A mass attached to a spring executes sim- ple harmonic motion in a horizontal plane with an amplitude of 3 . 64 m. At a point 1 . 3468 m away from the equilibrium, the mass has speed 0 . 886 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 . Correct answer: 23 . 9816 s. Explanation: Let A = 3 . 64 m , x = 1 . 3468 m , and v = 0 . 886 m / s . The simplest solution uses the equation for 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 , so parenleftBig x A parenrightBig 2 + parenleftBig v A parenrightBig 2 = 1 , and parenleftBig v parenrightBig 2 = A 2 x 2 . Consequently, = v A 2 x 2 = . 886 m / s radicalbig (3 . 64 m) 2 (1 . 3468 m) 2 = 0 . 262 s 1 and T = 2 = 2 . 262 s 1 = 23 . 9816 s . 003 (part 1 of 2) 10.0 points A 1 . 91 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4 . 59 N / m. The object is displaced 4 . 4 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the magnitude of the force acting on the object 3 . 67 s after it is released? Correct answer: 16 . 7374 N. Explanation: salazar (ggs332) HW11 TSOI (58160) 2 Let : A = 4 . 4 m , k = 4 . 59 N / m , m = 1 . 91 kg , and t = 3 . 67 s . The block exhibits simple harmonic motion, so the displacement is x = A cos( t ) where the angular frequency is = radicalbigg k m . Thus the force is F = k x = k A cos parenleftBigg radicalbigg k m t parenrightBigg = (4 . 59 N / m) (4 . 4 m) cos bracketleftBigg radicalBigg 4 . 59 N / m 1 . 91 kg (3 . 67 s) bracketrightBigg = 16 . 7374 N a force of 16 . 7374 N directed to the left....
View Full Document

Page1 / 11

HW11 Solution - salazar (ggs332) HW11 TSOI (58160) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online