A machine part is vibrating along the x-axis in simple harmonic motion with a period of 0,29 s and a range (from the maximum in one direction to the maximum in the other) of 4,8 cm. At time t = 0 it is at its central position and moving in the +x direction. What is its position (cm) when t = 49 s? Answer in two decimal places.
Solution)
We know,
x = A sin(ωt)
Hete, amplitude,
A = (4.8 cm) / 2
A = 2.4 cm
Angular velocity,
ω = 2π / T
ω = 2π / (0.29 s)
Also, at time t = 49 s,
x = (2.4 cm) sin( [2π / (0.29 s)] × (49 s) )
x = -0.775 cm
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