mathman2428
14.04.2021 •
Physics
a 1 kg block oscillates on a spring with spring constant 20N/m at t=0s the block is 20 cm to the right of the equilibrium position and movin to the left at a speed of 100 cm/s determine the period of oscillation and draw a graph of position versus time
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Ответ:
T = π/√5 ≈ 1.40s
Explanation:
ω = √(k/m) = √(20/1) = 2√5 rad/s
T = 2π/ω = 2π/2√5 = π/√5 ≈ 1.4049 s
x(t) = Acos(ωt + φ)
v(t) = -Aωsin(ωt + φ)
v(t)/x(t) = -ωtan(ωt + φ)
If we let RIGHT be the positive direction and evaluate at t = 0
-0.100 / 0.20 = -2√5tan(φ)
φ = arctan(-0.5/-2√5) = 0.111341...≈ 0.111 rad
0.20 = Acos(0.111)
A = 0.201246...≈ 0.201 m
x(t) = 0.201cos(2√5t + 0.111)
Ответ:
1st one's true that's all I know.