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iwonde
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[SOLVED] Simple Harmonic Motion
A cheerleader waves her pom-pom in simple harmonic motion with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach, and which ones cannot? Explain.
abs(a_max)= ω^2A, A = amplitude
abs(v_max)=ωA
ω=2πf, f=frequency
a = -ω^2 sqrt(A^2 - x^2)
(a) ω=2π(0.85)= 5.34 rad/s
abs(a_max)= (5.34)^2(0.18)= 5.13m/s^2
abs(v_max)= (5.34)(0.18) = 0.961 m/s
(b) a = -(5.34)^2 sqrt((0.18)^2 - (0.09)^2) = -4.45m/s^2
speed = abs(v) = (5.34) sqrt((0.18)^2 - (0.09)^2) = 0.832 m/s
I don't know how to approach (c) and (d).
Homework Statement
A cheerleader waves her pom-pom in simple harmonic motion with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach, and which ones cannot? Explain.
Homework Equations
abs(a_max)= ω^2A, A = amplitude
abs(v_max)=ωA
ω=2πf, f=frequency
a = -ω^2 sqrt(A^2 - x^2)
The Attempt at a Solution
(a) ω=2π(0.85)= 5.34 rad/s
abs(a_max)= (5.34)^2(0.18)= 5.13m/s^2
abs(v_max)= (5.34)(0.18) = 0.961 m/s
(b) a = -(5.34)^2 sqrt((0.18)^2 - (0.09)^2) = -4.45m/s^2
speed = abs(v) = (5.34) sqrt((0.18)^2 - (0.09)^2) = 0.832 m/s
I don't know how to approach (c) and (d).