A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?

The time taken for the spring to reach new equilibrium position is 0.26 s.

The given parameters;

The general wave equation is given as;

The angular frequency is given as follows;

The spring constant is calculated as;

The angular frequency is calculated as follows;

The time taken for the spring to reach new equilibrium position is calculated as follows;

Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.

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The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

(1)

Where:

- Position of the mass as a function of time, measured in meters.

- Amplitude, measured in meters.

- Spring constant, measured in newtons per meter.

- Mass of the block, measured in kilograms.

- Time, measured in seconds.

- Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

(2)

Where:

- Gravitational acceleration, measured in meters per square second.

- Deformation of the spring due to gravity, measured in meters.

If we know that , and , then the spring constant is:

If we know that , , , and , then (1) is reduced into this form:

(1)

And now we solve for . Given that cosine is a periodic function, we are only interested in the least value of such that mass reaches equilibrium position. Then:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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