A 208g oscillator has a speed of 112cm/s when its displacement is 1.60cm and 48.5cm/s when its displacement is 5.70cm .What is the oscillator's maximum speed?

Question

alekos · Answer

is there more info to this question? i don't think there's enough to go on

anonymous · Answer

For a linear osscilator F = k*x

dF/dt = k (dx/dt)

Force = 0 at x=0
      = maximum at x=maximum

Force varies from zero to a maximum in a linear fashion so the velocity varies from zero to a maximum in a linear fashion.

We need to find the parameters for velocity = slope*(displacement) + intercept

y = mx + b

Change between the two given points is

velocity = (48.5-112)
         = -63.5 cm/s

distance = (5.7-1.6)
         = 4.1 cm

so slope = -(63.5/4.1)
         = -15.48 (cm/s)/(cm)

y = (-15.48)*x + b

using 112 = (-15.48)*1.6 + b gives

b = 112 + (15.48*1.6)
  = 136.8

Maximum speed occurs at x=0 (the y-intercept)

y = (-15.48)*0 + 136.8
  = 136.8 cm/s

Maximum displacement occurs at y = 0 = (-15.48)*x + 136.8

x = (136.8/15.48)
  = 8.83 cm

alekos · Answer

the formula your using is incorrect
For displacement x =  $$Asin  \omega$$ t  where A is the max displacement $$\omega $$ is the frequency and t is the time
For velocity dx/dt = $$A \omega \cos \omega t$$