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ID: 1453211 • Letter: E

Question

ei WileyPLUS dugenwileyplus.com/edugen/student/mainfr.uni edugen wileyplus.com/edugen/shared/assignment/test/qprint.uni rint by: AHMED ALFAIFI HYS 1303 001: Introductory Mechanics/Copy of Chapter 15 Homework Problems Chapter 15, Problem 014 A simple harmonic oscillator consists of a block of mass 1.60 ko attached to a spning of spring constant 440 N/m. when t- 0.810 s, the position and velocity of the block are x- 0.140 m and v-4.300 m/s. (a) what is the amplitude of the oscilations? what were the (b) position and (c) velocity of the block att- s? (a) Number[D Cb) Number }‘Units [ - Units (c) Number Question Atteniptsi o uf G used

Explanation / Answer


angular velocity w = sqrt(k/m) = sqrt(440/1.6) = 16.6 rad/s

at t = 0.81 sec

x = A*cos(w*t)

A = x/(cos(w*t)) = 0.14/(cos(16.6*0.81)) =0.219 m

v = 4.3 m/s

a) A = 0.219 m

B) position at t = 0 sec is

x = 0.219*cos(w*t) = 0.219A

v = 0 m/s