1 year ago

The force on the wire is directed?

Physics

1 answer:

USPshnik [31]1 year ago

5 0

Answer: A

Explanation:

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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

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Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

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Hope this helps! ^^

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