oblem Chapter 15, Problem at t correct. Try again. 0 s? 014 block is 0.453 answe

Question

oblem Chapter 15, Problem at t correct. Try again. 0 s? 014 block is 0.453 answer partially velocity of the Your (a) Nut Mail thlg har Outlook FULL SCREEN PRINTER VERSION BACK NEXT A simple harmonic oscillator consists of a block of mass 1.20 kg attached to a spring of spring constant 120 N/m. When t 1.10 s, the position and velocity of the block are x w 0.148 m and v- 4.280 m/s. (a) What is the amplitude of the oscillations? What were the (b) postition and (c) (b) Number 0.0714 Units (c) Number 1.461 Click if you would like to show Work for this question open show Work Question Attempts: 2 of 5 used SAVEFORLATER SUBMTTANSwER nghts reserved. Copyright zoob 2016 by Donn wiley B sons, Inc or related corpanies. Aii sannement envacy Policy I R2000 2016 2ghn sens insa All Rights Reserved. A Divison of hehn MilevAsens. Version 4.21.15 MacBook Air

Explanation / Answer

m w^2 = k

w = sqrt(120 / 1.20) = 10 rad/s

x = A sin(wt + phi)

v = dx/dt = A w cos(wt + phi)

at t = 1.10 s

0.148 = A sin(wt + phi) .......(i)

4.280 = (A x 10) cos(wt + phi)

0.428 = A cos(wt + phi) .....(ii)

(i) / (ii) =>

tan(wt + phi) = 0.346

w t + phi = 0.333 rad

10 x 1.10 + phi = 0.333

phi = -10.67 rad

0.148 = A sin(0.333 rad)

A = 0.453 m .......Ans(A)

(B) at t = 0

x = 0.453 sin(-10.67) = 0.429 m

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