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## (Figure 1) is the velocity-versus-time graph of a particle in simple harmonic motion.?

A/

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

gives us

amplitude A = 115 cm = 1.15 m ◄

B/

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

C/

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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