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A spring has a natural length of 0.75 m and a 5-kg mass. A force of 25 N is needed to keep the spring stretched to a length of 1 m. By Hooke's Law, the force required to stretch a spring by length \( x \) is \( kx \), where \( k \) is the spring constant.

If the spring is released after being stretched to a length of 1 m (i.e., the initial velocity is 0), what is the ODE that governs the position (or elongation) \( x(t) \) as a function of time?

A. \(* + 20x = 0\)
B. \(Ï + 5x = 0\)
C. \(3 + 100.2 = 0\)
D. \(+ 20x = 0\)
E. \(C - 20x = 0\)
F. None

Solve the ODE in the preceding problem to find the position \( x(t) \) as a function of time. Remember, it is an Initial Value Problem: \( x(0) \) is the initial elongation and \( i(0) \) is the initial velocity.

A. \(\cos(\sqrt{20}t)\)
B. \(0.25 \cos(\sqrt{20}t)\)
C. \(\cos(\sqrt{5}t)\)
D. \(0.25 \cos(10+)\)
E. \(0.75 \cos(\sqrt{5}t)\)
F. \(0.125e^{\sqrt{20}t} + 0.125e^{-\sqrt{20}t}\)

Answer :

Final answer:

The ODE that governs the position x(t) as a function of time when the spring is released after stretching to a length of 1 m is x''(t) = -k/5.

Explanation:

To find the ODE that governs the position x(t) as a function of time when the spring is released, we can use Hooke's law and Newton's second law of motion.

Hooke's law states that the force required to stretch or compress a spring is proportional to the displacement from its natural length. Mathematically, this can be expressed as:

F = -kx

Where F is the force applied to the spring, k is the spring constant, and x is the displacement from the natural length.

Newton's second law of motion states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this can be expressed as:

F = ma

Where F is the force applied to the object, m is the mass of the object, and a is the acceleration of the object.

In this case, the force applied to the spring is given as 25 N, the mass of the spring is 5 kg, and the displacement from the natural length is 1 m. We can equate the force from Hooke's law to the force from Newton's second law:

-kx = ma

Substituting the given values:

-k(1) = 5a

Simplifying the equation:

-k = 5a

Dividing both sides by 5:

-k/5 = a

Therefore, the ODE that governs the position x(t) as a function of time when the spring is released is:

x''(t) = -k/5

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Rewritten by : Jeany

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