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maksim [4K]
4 years ago
5

A 2.00 kg object is attached to a spring and placed on frictionless, horizontal surface. Ahorizontal force of 18.0 N is required

to hold the object at rest when it is pulled 0.250 mfrom its equilibrium position. What is the angular frequency, in rad/s, of the oscillation?
Physics
1 answer:
velikii [3]4 years ago
7 0

Answer:

6 rad/s

Explanation:

In a spring the angular frequency is calculated as follows:

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

where \omega is the angular frequency, m is the mass of the object in this case m=2kg, and k is the constant of the spring.

To calculate the angular frequency, first we need to find the constant k which is calculated as follows:

k=\frac{F}{x}

Where F is the force: F=18N, and x is the distance from the equilibrium position: x=0.25m.

Thus the spring constant:

k=\frac{18N}{0.25m}

k=72N/m

And now we do have everything necessary to calculate the angular frequency:

\omega=\sqrt{\frac{k}{m} }=\sqrt{\frac{72N/m}{2kg} }=\sqrt{36}

\omega=6rad/s

the angular frequency of the oscillation is 6 rad/s

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Given the data in the question;

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