Question

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 1

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.