Question

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A particle executes linear SHM with frequency 0.11 Hz about the point x = 0. At t = 0, it has displacement x = 0.33 cm and zero velocity. For the motion, determine the

(a) displacement at t = 2.2 s, and (b) velocity at t = 2.2 s.

Answer 1

Let the equation of motion of the particle be given by:

x=A sin(t + ) where x is position, A is amplitude, is angular frequency, t is time, is initial phase.

=2f, where f is frequency.

Here,f=0.11 Hz,so, =2*0.11=0.22

v=(A²-x²)^1/2 , where v is speed,   is angular frequency,A is amplitude,x is position

Here,at x=0.33 cm,v=0 cm/s=>0=0.22(A²-0.33²)^1/2=>A²=0.33²=>A=0.33 cm (As amplitude cannot be negative)

Initial phase = sin^-1(x₀/A) where x₀ is initial position and A is amplitude.

So,here,=sin^-1(0.33/0.33)=/2

So,equation of motion is given by:

x=0.33sin(0.22t+/2) cm

So,at t=2.2 s, displacement=0.33sin(0.22*2.2+/2_)=0.33 sin(0.984)=0.0178 cm=1.78 * 10^-2 cm

v=d(x)/dt, where v is velocity, x ts position as a function of time and t is time.

So,v=0.33d[ sin(0.22t+/2) ]/dt=0.33*0.22 cos(0.22t+/2) cm/s=0.228 cos(0.22t+/2) cm/s

If t=2.2 s, v=0.228 cos(0.22(2.2)+/2) cm/s=0.228 cos(0.984) cm/s=0.228 cm/s