An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.1 rad/s. Calculate the distance d by which the spring stretches from its unstrained length when the object is allowed to hang stationary from it.

To calculate the distance d by which the spring stretches from its unstrained length when the object is allowed to hang stationary from it, we can use Hooke's Law.

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the object from its equilibrium position. Mathematically, it can be written as:

F = -k * x

Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement from the equilibrium position.

In this case, the object is stationary, so the net force on the object is zero. Therefore, we can set the force due to gravity equal to the force exerted by the spring.

mg = k * d

Where:
m is the mass of the object,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
d is the distance by which the spring stretches.

To solve for d, we can rearrange the equation:

d = mg / k

Given that the angular frequency ω is 25.1 rad/s, we can relate it to the spring constant k using the equation:

ω = sqrt(k / m)

Solving for k, we have:

k = ω^2 * m

Substituting this expression for k into the equation for d, we get:

d = mg / (ω^2 * m)

The mass m cancels out, and we can plug in the values to calculate d.

18m