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hammer [34]
3 years ago
6

A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially

released from the equilibrium position with an upward velocity of 8 m/s. Find the equation of motion. x(t)
Physics
1 answer:
zysi [14]3 years ago
4 0

Answer:

x(t) = -8sin2t

Explanation:

See the attachment for solution

From my solving, we can deduce that w² = 4, and thus, w = 2

Therefore, the general solution is

x(t) = c1 cos2t + c2 sin2t

Considering the final variable, we can conclude that

x(0) = 0

x'(0) = -8 m/s

The final solution, thus

x(t) = -8sin2t

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Answer:

The potential difference between points A and B is 278.95 volts.

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Given :

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V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }

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V₃ = -363.15 V

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