A person bounces up and down on a
trampoline, while always staying in contact with it. The motion is
simple harmonic motion, and it takes 2.73 s to complete one cycle.
The height of each bounce above the equilibrium position is 45.2
cm. Determine (a) the amplitude and (b) the angular frequency of
the motion. (c) What is the maximum speed attained by the
person?
\(A=45.2 cm=.452 m\)
\(\omega=\frac{2\pi}{T}=\frac{2\pi}{2.73 s}=2.30 rad/s\)
\(v_{max}=A\omega=(.452 m)(2.30 rad/s)=1.04 m/s\)
A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic m...
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An object with mass 3.9 kg is executing simple harmonic
motoon, attached to a spring with spring constant 250 N/m. When the
object is 0.018 m from its equilibrium position, it is moving with
a speed of 0.50 m/s.
A) Calculate the amplitude of the motion
B) Calculate the maximum speed attained by the object
Thank you!
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