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(Figure 1) is the velocity-versus-time graph of a particle in simple harmonic motion.?
We observe on the graph one complete cycle, so
the period T = 12 s, and so the
angular frequency ω = 2π/T = 0.52 rad/s
We also observe from the graph that the maximum velocity is 60 cm/s
We know (or ought to know) that
max velocity = amplitude * angular frequency, so
60 cm /s = A * 0.52rad/s
amplitude A = 115 cm = 1.15 m ◄
I hate these phase angle questions, since it depends on how you model the motion. If we use
Vx = A*cos(2πt/12 + φ), then φ = π/3 radians. ◄
But if you use Vx = A*cos(2πt/12 – φ), then φ = -π/3 radians.
And you might even use sin instead of cos, in which case φ = ± 5π/6 rads
If Vx = 60cm/s*cos(2πt/12 + π/3), then
x = 1.15m * sin(2πt/12 + π/3)
and so when t = 0,
x = 1.15m * sin(π/3) = 0.99 m ◄
and this should be true regardless of how you model the motion.
Hope this helps!
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