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 2.82 s to complete one cycle. The height of each bounce above the equilibrium position is 40.8 cm. Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?
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(a) .408 m
(b) The equation I have is f=1/T. So I thought that it would be 1/(2.82)= .355, but apparently that is not the correct answer. Can you help please?
(c) I cant answer it without part (b).
You need to take the number you get from that equation and then plug it into w=2pif
w=2pi(.355) and that will give you the correct answer for angular frequency
I will try again without rounding. How many decimal places should I put?
Sure, let's work through this problem step by step.
(a) The amplitude of the motion is the maximum height of each bounce above the equilibrium position. In this case, it is given as 40.8 cm. To convert this to meters, divide by 100: 40.8 cm = 0.408 m. So, the amplitude of the motion is 0.408 m.
(b) The angular frequency (ω) is related to the time period (T) by the formula ω = 2π/T. Given that the time period is 2.82 s, we can substitute it into the formula: ω = 2π/2.82 ≈ 2.226 rad/s. So, the angular frequency of the motion is approximately 2.226 rad/s.
(c) To determine the maximum speed attained by the person, we can use the formula v_max = ω * A, where v_max represents the maximum speed, ω is the angular frequency, and A is the amplitude. Based on the values given in parts (a) and (b), we can substitute them into the formula: v_max = 2.226 * 0.408 ≈ 0.907 m/s. So, the maximum speed attained by the person is approximately 0.907 m/s.
So, the answers are:
(a) The amplitude is 0.408 m.
(b) The angular frequency is approximately 2.226 rad/s.
(c) The maximum speed attained by the person is approximately 0.907 m/s.