9. A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position,and it subsequently undergoes simple harmonic oscillations.Find

a)the force constant of the spring
b)the frequency of the oscillations

To find the force constant of the spring (a), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

Hooke's Law can be mathematically stated as:

F = -kx

where F is the force exerted by the spring, k is the force constant (also known as the spring constant), and x is the displacement from the equilibrium position.

In this case, we are given the force required to hold the object at rest when it is pulled 0.200 m from its equilibrium position. Therefore, we can use this information to calculate the force constant of the spring.

Given:
Force (F) = 20.0 N
Displacement (x) = 0.200 m

Using Hooke's Law, substituting the given values, we have:

20.0 N = -k * 0.200 m

To find the force constant (k), we rearrange the equation:

k = -20.0 N / 0.200 m

Calculating, we get:

k = -100 N/m

Since the negative sign indicates that the force is directed opposite to the displacement, we drop the negative sign. Therefore, the force constant (spring constant) of the spring is 100 N/m.

To find the frequency of the oscillations (b), we can use the formula:

f = 1 / (2π) * √(k/m)

where f is the frequency, k is the force constant, and m is the mass of the object.

Given:
Force constant (k) = 100 N/m
Mass (m) = 2.00 kg

Substituting these values into the formula, we have:

f = 1 / (2π) * √(100 N/m / 2.00 kg)

Calculating, we get:

f ≈ 1.59 Hz

Therefore, the frequency of the oscillations is approximately 1.59 Hz.