Part A A0.830-kg block oscillates on the end of a spring whose spring constant i

Question

Part A A0.830-kg block oscillates on the end of a spring whose spring constant is k 40.0 N/m . The mass moves in a fluid which offers a resistive force What is the period of the motion? F-bu, where b 0.620 N s/m Submit Request Answer Part B What is the fractional decrease in amplitude per cycle? Submit Request Answer Part C Write the displacement as a function of time t in seconds if at t0, 0, and att1.00s, 0.120m. ¢-(0.289m) sin(6.93rad/s)t] (0.289m)e-07* sin[(6.93rad/s)t] (0.289m) cos (6.93rad/s)t) O z-(0.289m)e-(0.3738-ytCOS(6.93rad/s)t] Submit Request Answer Provide Feedback Next >

Explanation / Answer

a)

Angular frequency of oscillation

W=sqrt[ (K/m) -(b/2m)2]

W=sqrt [ (40/0.83) -(0.62/2*0.83)2]

W=6.932 rad/s

The Period of the motion

T=2pi/W =2pi/6.932 =0.9064 s

b)

dA/A =[e-bt/2m-e-b(t+T)/2m]/e-bt/2m =1-e-bT/2m

dA/A =1-e-0.62*0.9064/2*0.83

dA/A =0.287

c)

DIsplacement of the wave

X(t) =Ae-bt/2mCos(Wt+phi)

at t=0 ,X=0

0=ACos(phi)

phi =Cos-1(0)=pi/2

at t=1 ,x=0.12 m

=>0.12 =Ae-0.62*1/2*0.83Cos(pi/2 +6.932*1)

A=0.2885 m =0.289 m

Therefore

X=(0.289 m)e-0.62*t/2*0.83Cos(6.932t +pi/2)

X=(0.289 m)e-(0.373 s^-1)sin[(6.93 rad/s)t]

Therefore answer is B

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