Question

A particle moves in simple harmonic motion with a frequency of 3.00 Hz and an amplitude of 3.30 cm. (a) Through what total distance does the particle move during one cycle of its motion? cm (b) What is it maximum speed? cm/sWhere does this maximum speed occur? as the particle passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these (c) Find the maximum acceleration of the particle. m/s2Where in the motion does the maximum acceleration occur? as the particle passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these

Answer 1

a) d= 13.2cm

b ) vmax =62.23cm/s

c) amax=1172.81cm/s2

Step-by-step explanation:

Given:

frequency is 3.00Hz,

the amplitude is 3.30cm.

(a)The total distance travelled by the particle during one cycle is.

d = 4A

d= 4×3.30cm

d = 13.2cm

b) The maximum speed of the particle is calculated as using the formula;

vmax = ωA

= 2πfA

= 2π (3.30cm/s)*(3.0cm/s)

=2× 3.147 × 3.30 × 3.0

vmax =62.23cm/s

The maximum speed occur at mean position(x,0) which means it has maximum speed when it passes through the equilibrium position.

(c) The maximum acceleration of the particle is

calculated below;

amax = ω2A

= (2πf)2A

= (2× 3.142×3)^2×3.30

amax=1172.81cm/s2

The particle has maximum acceleration at the turning points.