PH101 HW Waves

PH101 HW Waves - This print-out should have 23 questions Multiple-choice questions may continue on the next column or page nd all choices before

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

View Full Document Right Arrow Icon
This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Apa r t ic leo sc i l la te supanddownins imp le harmonic motion. Its height y as a Function oF time t is shown in the diagram. 1 2 3 4 5 5 5 y (cm) t (s) At what time t in the period shown does the particle achieve its maximum positive ac- celeration? 1. None oF these; the acceleration is con- stant. 2. t =3s 3. t =1s correct 4. t =4s 5. t =2s Explanation: This oscillation is described by y ( t )= - sin ± πt 2 ² , v ( t )= dy dt = - π 2 cos ± πt 2 ² a ( t )= d 2 y dt 2 = ³ π 2 ´ 2 sin ± πt 2 ² . The maximum acceleration will occur when sin ± πt 2 ² =1,orat t =1s . ±rom a non-calculus perspective, the veloc- ity is negative just beFore t =1ss incethe particle is slowing down. At t =1s ,thepar- ticle is momentarily at rest and v =0 . Just aFter t =1s,thev e loc i tyi spo s i t iv es ince the particle is speeding up. Remember that a = Δ v Δ t , acceleration is a positive maximum because the velocity is changing From a nega- tive to a positive value. 002 10.0 points A128Nobjectv ibratesw ithaper iodoF4 .76 swhenhangingFromaspring. What is the spring constant oF the spring? The acceleration oF gravity is 9 . 81 m / s 2 . Correct answer: 22 . 7346 N / m. Explanation: Let : F g =128N , T =4 . 76 s , and g =9 . 81 m / s 2 . T =2 π µ m k ± T 2 π ² 2 = m k k = m ± 2 π T ² 2 = F g g ± 2 π T ² 2 = 128 N 9 . 81 m / s 2 ± 2 π 4 . 76 s ²
Background image of page 1

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

View Full DocumentRight Arrow Icon
Explanation: Let : m =52 . 2g=0 . 0522 kg , k =14 . 8N / m , and A =25 . 1cm=0 . 251 m . The speed is v = ± k m ( A 2 - x 2 ) = ² 14 . 8N / m 0 . 0522 kg [(0 . 251 m) 2 - (0 . 1255 m) 2 ] = 3 . 66016 m / s . 004 (part 1 of 3) 10.0 points A541gmassisconnectedtoal ightspr ing of force constant 3 N / mthatisfreetooscillate on a horizontal, frictionless track. The mass
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/03/2010 for the course PH 101 taught by Professor Jones,s during the Spring '08 term at Alabama.

Page1 / 7

PH101 HW Waves - This print-out should have 23 questions Multiple-choice questions may continue on the next column or page nd all choices before

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