How do you find the centre of motion of a particle moving in simple harmonic motion of period 8hrs and amplitude 6m. When t=3hrs and x=4m?

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Answer 1 · 2018-01-31T10:59:52

The center is #=-0.055rad#

The center of motion of a simple harmonic motio is when the velocity is maximum.

The amplitude is #a=6m#

The period is #T=8h#

The angular velocity is #omega=(2pi)/T=(2pi)/8=pi/4=0.785radh^-1#

The maximum velocity is

#v_m=axxomega=6*0.785=4.71mh^-1#

When #t=3h# and #x=4m#

#x=asin(omegat+phi)#

#4=6sin(0.785*3+phi)#

#sin(0.785*3+phi)=2/3#

#0.785*3+phi=0.73#

#phi=0.73-0.785=-0.055rad#

Therefore,

#x=6sin(0.785t-0.055)#

The phase of oscillation is #phi=-0.055#

graph{6sin(0.785x-0.055) [-5.55, 6.934, -1.604, 4.64]}