ctice Assignment Gradebook ORION Downloadable eTextbook ment El FULL SCREEN PRIN
ID: 2041108 • Letter: C
Question
ctice Assignment Gradebook ORION Downloadable eTextbook ment El FULL SCREEN PRINTER VERSION BACK NEXT Chapter 15, Problem 014 A simple harmonic oscillator consists of a block of mass 4.90 kg attached to a spring of spring constant 130 N/m. Whent 2.20 s, the position and velocity of the block are x-0.165 m and v = 3.170 m/s (a) what is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t 0 s? (a) Number (b) Number (c) Number Units Units SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM MATH HELP SAVE FOR LATER SUBMIT ANSWER Question Attempts: UnlimitedExplanation / Answer
given
m = 4.90 kg
k = 130 N/m
at x = 0.165 m, v = 3.17 m/s
a) let A is the amplitude of motion.
Apply conservation of energy
total energy = sum of potential and kinetic energy
(1/2)*k*A^2 = (1/2)*k*x^2 + (1/2)*m*v^2
A^2 = x^2 + (m/k)*v^2
A = sqrt(x^2 + (m/k)*v^2)
= sqrt(0.165^2 + (4.9/130)*3.17^2)
= 0.637 m <<<<<<<<<<<----------------------Answer
b) angular frequency of motion,
w = sqrt(k/m)
= sqrt(130/4.9)
= 5.15 rad/s
let phi is the phase constant.
x = A*cos(w*t + phi)
at t = 2.2 s, x = 0.165 m
so,
0.165 = 0.637*cos(5.15*2.2 + phi)
0.165/0.637 = cos(11.33 + phi)
0.259 = cos(11.33 + phi)
11.33 + phi = cos^-(0.259)
11.3 + phi = 1.309
phi = 1.409 - 11.33
= -9.92 radians
so,
at t = 0,
x = A*cos(w*t + phi)
= 0.637*cos(0 - 9.92)
= -0.560 m <<<<<<<<-----------------Answer
c) v = dx/dt
= A*(-sin(wt + phi)*w
= -A*w*sin(wt + phi)
at t = 0,
v = -0.637*5.15*sin(0 - 9.92)
= -1.56 m/s <<<<<<<<-----------------------Answer
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