A body oscillates with simple harmonic motion accordingto the

equation:

x = (6.0 m) cos [ (3pai rad/s) t + pai/3 rad]

at t = 2.0 s, what are:

a.) the displacement

b.) the velocity

c.) the acceleration

d.) the phase of the motion

Also, what are

e.) the frequency and

f.) the period of the motion?

Answer: (a) displacement = 6.0m

(b) velocity = 8.996m/s

(c) acceleration

(d) phase of motion = 2.999

(e) frequency = 1.499hz

(f) period of motion = 0.667sec

Explanation:Given that,

X = (6.0m) cos (3πrad/s) t + π/3rad ... (i)

Compare equation (i) to wave equation,

Y = A cos (wt) + kx ... (ii)

(a) displacement which is also known as amplitude A. So by comparing equation (i) & (ii), we have A = 6.0m

Therefore, displacement = 6.0m

(b) velocity=B/T (where B = wavelength and T=period)

Wavelength B=2π/k (where k=π/3, comparing the two equations)

Therefore, wavelength B=2π:π/3=6

Period T = 2π/w (where w=3π, comparing the two equations)

Therefore, period T=2π:3π=0.667sec

Therefore, Velocity V=B/T=6/0.667=8.996m/s

(d) phase of motion=t/T (where t = elapsed time and T=period)

T=0.667sec (from (c) solving for velocity)

t=2.0s, as given in the question

Therefore, phase of motion = t/T=2/0.667=2.999

(e) frequency=1/T (where T is the period)

T=0.667secs (recall from (c) while solving for velocity)

Therefore, frequency=1/T=1/0.667=1.499hz

(f) period=1/f (where f = frequency)

Therefore, period=1/f=1/1.499=0.667secs.