Question

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the maximum value, at t = 0, moving to the right. The amplitude of the motion is 2.00 cm and the frequency is 1.50 Hz. (a) Find an expression for the position of the particle as a function of time. Determine (b) the maximum speed of the particle and (c) the earliest time (t > 0) at which the particle has this speed. Find (d) the maximum positive acceleration of the particle and (e) the earliest time ( t > 0) at which the particle has this acceleration. (f) Find the total distance traveled by the particle betweent= 0 and t= 1.00 s.

Answer 1

f = 1.5
w = 2pi*f = 3pi
x = Asin(wt+ C)

A = 2.00 cm
C = 0 as it starts from origin at t = 0

x = 2 sin(3pi*t) cm

v = dx/dt = 18.84cos(3pi*t) cm/s

a = dv/dt = - 1.77sin(3pi*t) m/s^2

so

a) x = 2sin(3pi*t) cm

b) max (v) => cos(3pi*t) = 1

Vmax = 18.84 cm/s

c) cos(3pi*t) = 1

for earliest time
3pi*t = pi (it cant be 0 as its at rest at t = 0 )

t = 1/3 = 0.33 seconds

d) for max positive acc

sin(3pi*t) = -1
a max = 1.77 m/s^2

e) sin(3pi*t) = -1

for earliest time
3 pi * t = 3pi/2

t = 1/2

t = 0.5 seconds

f) for the total distance traveled between t = 0 and t = 1 s

X = 4*2 + 2*2 = 12 cm