A cheerleader waves her pom-pom in SHM with an amplitude of 17.0 cm and a freque

Question

A cheerleader waves her pom-pom in SHM with an amplitude of 17.0 cm and a frequency of 0.900 Hz .

Part A

Find the maximum magnitude of the acceleration.

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Part B

Find the maximum magnitude of the velocity.

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Part C

Find the acceleration when the pom-pom's coordinate is x= 9.50 cm .

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Part D

Find the speed when the pom-pom's coordinate is x= 9.50 cm .

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Part E

Find the time required to move from the equilibrium position directly to a point a distance 12.8 cm away.

A cheerleader waves her pom-pom in SHM with an amplitude of 17.0 cm and a frequency of 0.900 Hz .

Part A

Find the maximum magnitude of the acceleration.

amax =   m/s2  

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Part B

Find the maximum magnitude of the velocity.

vmax =   m/s  

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Part C

Find the acceleration when the pom-pom's coordinate is x= 9.50 cm .

a =   m/s2  

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Part D

Find the speed when the pom-pom's coordinate is x= 9.50 cm .

v =   m/s  

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Part E

Find the time required to move from the equilibrium position directly to a point a distance 12.8 cm away.

t =   s  

Explanation / Answer

(a)Frequency = 0.9 Hz

w = 2*pi*0.9 = 5.655 rad/s

amax = A*w^2 = 0.17* (5.655)^2 = 5.44 m/s^2

(b) vmax = A*w = 0.17*5.655 = 0.96135 m/s

(c) x = A sin(wt) = 0.095

Hence acceleration = -A*w^2*sin(wt) = -0.095*w^2 = -3.04 m/s^2

(d) Now

x = A sin(wt) = 0.095

sin(wt) = 0.559

cos(wt) = ( 1- sin(wt)^2)^0.5

= 0.829

Hence speed = A*w*cos(wt) = 0.17*5.655*0.856 = 0.797 m/s

(e) Now

x = Asin(wt) = 0.128

sin(wt) = 0.753

(wt) = 0.853

t = 0.151 s

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