A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. The frequency of the motion is 0.25 Hz. Determine a) the period, (b) the angular frequency, (c) the amplitude, (d) the displacement at time t, (e) the velocity at time t, (f) the maximum speed, (g) the maximum acceleration,

(a) The period is 4s

(b) The angular frequency is pi/2 radians

(c) The amplitude is 0.37cm.

(d) The displacement at time is  (0.37 cm) cos((pi/2)*t)

(e) The Velocity at time t is v = (0.58 cm)(sin((pi/2)*t)

(f) The maximum speed is  

(g) The  maximum acceleration is 0.91 cm/s^2

Explanation:

We have a particle which oscillates with frequency of f = 0.25 Hz about the point x = 0.At t = 0, the displacement of the particle is = 0.37 cm and its velocity is zero.

(a) The period of the oscillations is,

so

T = 1/(0.25 Hz)

T = 4.0s

(b) The angular frequency is,

f = = 2()(0.25 Hz) = radians

(c) Since

The amplitude is the maximum displacement that the particle makes from the equilibrium point, or when the speed of the particle is zero,

that is

= 0.37 cm

(d) The displacement as a function of t is given be,

x = cos(ωt+Φ)

as x =  t = 0, we get cos(Ф) = 1 = 0

so this equation becomes

x= (0.37 cm) cos((pi/2)*t)

(e) Now we need to find the speed of the particle as a function of t

the speed is the derivative of the displacement that is

= -(0.37)(pi/2)(sin((pi/2)*t)

so the velocity at time t is

 v = (0.58 cm)(sin((pi/2)*t)

(f) Since

sin(ωt+Ф)

then from part (e) we get

(g)

The amplitude of the maximum acceleration is

ω^2

      = (0.37 cm) (pi/2) = 0.91 cm/s^2

this is the maximum acceleration

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