A particle moving in simple harmonic motion with a period of 1.5 s passes throug

Question

A particle moving in simple harmonic motion with a period of 1.5 s passes through the equilibrium point at time t(sub o)=0 with a speed of 1.00 m/s to the right. At a time later, the particle is observed to move to the left with a speed of 0.50 m/s.

a) Write the equations, as a function of time, for the

- displacement of this particle

- velocity of this particle

- acceleration of this particle

b) Calculate

- frequency of this motion

- the smallest value of elapsed time t for the particle to be moving to the left with a velocity of 0.50 m/s

Explanation / Answer

a)

x=Asint

v=dx/dt=Acost

a=dv/dt= -^2Acost= -^2x

b)

=sqrt(k/m)

At t=0s , v=1.0m/s

Plug in velocity eqn,

1.0=(2/1.5)Acos[(2/1.5)*0]

A=0.2387m

Now plug value of A=0.2387 and v=0.50m/s the same eqn

0.50=(2/1.5)Acos[(2/1.5)*t]

1.25s

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