Question

A simple harmonic oscillator's position is given by

y(t) = (0.950 m)cos(11.8t − 6.15).

Find the oscillator's position, velocity, and acceleration at each of the following times. (Include the sign of the value in your answer.)

(a)    

t = 0

position	m
velocity	m/s
acceleration	m/s2

(b)    

t = 0.500 s

position	m
velocity	m/s
acceleration	m/s2

(c)    

t = 2.00 s

position	m
velocity	m/s
acceleration	m/s2

Answer 1

position, y(t) = (0.950 m)cos(11.8t − 6.15)

velocity, v(t)= dy/dt => v(t) = -(0.950 *11.8)sin(11.8t − 6.15).= -(11.21 m/s)sin(11.8t − 6.15).

accln a(t) = dv/dt = >a(t) = -(11.21*11.8 )cos(11.8t − 6.15).= -(132.278m/s^2)cos(11.8t − 6.15)

a)at t=0

position=y(0)=(0.950 m)cos(− 6.15) =0.942 m

velocity = v(0) = -(11.21 m/s)sin(− 6.15) = -1.49 m/s

accln = a(0)= -(132.278m/s² )cos(− 6.15) =-131.106 m/s² .

b) at t= 0.500s

position=y(0.500)=(0.950 m)cos(11.8*0.500− 6.15) =0.920 m

velocity = v(0.500) = -(11.21 m/s)sin(11.8*0.500− 6.15) = 2.77 m/s

accln = a(0.500)= -(132.278m/s² )cos(11.8*0.500− 6.15) =-128.165 m/s² .

c) at t=2.00 s

position=y(2.00)=(0.950 m)cos(11.8*2.00− 6.15) =0.162 m

velocity = v(2.00) = -(11.21 m/s)sin(11.8*2.00− 6.15) = 11.04 m/s

accln = a(2.00)= -(132.278m/s² )cos(11.8*2.00− 6.15) =--22.54 m/s² .

answered by: anonymous