An elastic band is hung on a hook and a mass is hung on the lower end of the ban

Question

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos t + 6 sin t, t ? 0, where s is measured in centimeters and t in seconds.

a.)Find the velocity and acceleration at time t.

c.)When does the mass pass through the equilibrium position for the first time? (Round your answer to two decimal places.)

d.)How far from its equilibrium position does the mass travel? (Round your answer to two decimal places.)

Explanation / Answer

A) When mass at equilibrium position, s = 0. So

2 cos(t) + 3 sin(t) = 0.
Sqrt(13) [sin(t + 0.588)] = 0
(t + 0.588) = 0, Pi, 2Pi, ...
t = 2.554s, 5.695s, where ignore negative value of -0.588s.

B) When mass at the most far distant from equilibrium (assume no friction), then ds/dt = v(t) = 0

Similar to above approach,
-2sin(t) + 3cos(t) = 0
Sqrt(13) [ sin( t - 0.983)] = 0
t - 0.983 = 0, Pi, 2Pi,...
t = 0.983, (Pi+0.983), (2Pi+0.983)

Sub t = 0.983 in s(t)
s(0.983) = 2 cos(0.983) + 3 sin(0.983)
s = 1.1094+2.4962 = 3.606cm

C) When v maximum, then dv/dt = a(t) = 0
-2cos(t) - 3sin(t) = 0
-[2cos(t) + 3sin(t)] = 0
-Sqrt(13) [sin(t + 0.588)] = 0
(t + 0.588) = 0, Pi, 2Pi, ...
t = 2.554s, 5.695s, where ignore negative value of -0.588s.

Sub t = 2.554s, 5.695s in s(t)

S=0

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