Question

A cheerleader waves her pom-pom in SHM with an amplitude of 18.8 cm and a frequency of 0.880 Hz .

a) Find the maximum magnitude of the acceleration.

b) Find the maximum magnitude of the velocity.

c) Find the acceleration when the pom-pom's coordinate is x= 8.70 cm .

d) Find the speed when the pom-pom's coordinate is x= 8.70 cm .

e) Find the time required to move from the equilibrium position directly to a point a distance 11.9 cm away.

Answer 1

(a)

Frequency = 0.88

W(ANGULAR FREQUENCY) = 2*PI*0.88 = 5.5292

ACCELERATIONmax = A*w^2 = 0.188* (5.5292)^2 = 5.7475 m/s^2

(b)

Maximum velocity = A*w = 0.188*5.7475 = 1.0805 m/s

(c)

x = A sin(wt) = 0.087

acceleration = -A*w^2*sin(wt) = -0.087*w^2 = -2.6597 m/s^2

(d)

x = A sin(wt) = 0.087

sin(wt) = 0.4627

cos(wt) = ( 1- sin(wt)^2)^1/2 = 0.8865

Speed = A*W*cos(wt) = 0.188*5.5292*0.8865 = 0.9215 m/s

(e)

x = Asin(wt) = 0.119

sin(wt) = 0.6329

(wt) = 0.6852

t = 0.1239 s