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Anna35 [415]
3 years ago
13

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

mpleted in 22.7 s. When this motion is viewed as a projection of circular motion, what are the radius, r, and angular velocity, ? , of the circular motion?
Physics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

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<h2>Answer:</h2>

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<h2>Explanation:</h2>

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Where;

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m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

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m₃ = 2kg

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x₂ = 3m

x₃ = 0m

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x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

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y = (6x0) + (4x0) + (2x3)) / 12

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