A force of 13 N is needed to keep a spring with a 2-kg mass stretched 0.25 m beyond its natural length. The damping constant of the spring is c = 8. If the mass starts at the equilibrium position with a velocity of 0.5 m/s, find its position at time t.

Position at time t = 0.25 m + 0.5 m/s * t - (13 N / (2 kg * 8)) * e^(-8t)

To find the position of the mass at time t, we can use the equation of motion for a damped harmonic oscillator:

m * x''(t) + c * x'(t) + k * x(t) = 0

Where:
m = mass (2 kg)
x(t) = position at time t
x'(t) = velocity at time t
x''(t) = acceleration at time t
c = damping constant (8)
k = spring constant (unknown)

First, let's find the value of k using the given force:

Force = k * x(t)

k = Force / x(t) = 13 N / 0.25 m = 52 N/m

Now we can set up the differential equation for the motion:

2 kg * x''(t) + 8 * x'(t) + 52 N/m * x(t) = 0

To solve this equation, we need to find the roots of the characteristic equation:

2r² + 8r + 52 = 0

Dividing the equation by 2 to simplify:

r² + 4r + 26 = 0

Using the quadratic formula, we find the roots:

r₁ = (-4 + √(4² - 4*1*26)) / 2 = -2 + √(-4)
r₂ = (-4 - √(4² - 4*1*26)) / 2 = -2 - √(-4)

Since the discriminant is negative, the roots are complex numbers. Let's denote them as α and β:

α = -2 + iω
β = -2 - iω

Where ω = √(-4)

The general solution for a damped harmonic oscillator with complex roots is:

x(t) = e^(-2t) * (A * cos(ωt) + B * sin(ωt))

To find the constants A and B, we can use the initial conditions:

When t = 0, x(0) = 0.25 m (the initial position)
When t = 0, x'(0) = 0.5 m/s (the initial velocity)

Plugging these values into the general solution:

x(0) = A * cos(0) + B * sin(0) = A
x'(0) = -2A * e^(-2*0) * cos(0) - 2B * e^(-2*0) * sin(0) + ω * (A * sin(0) - B * cos(0)) = -2A + ωB

So, we have two equations:

A = 0.25
-2A + ωB = 0.5

Substituting ω = √(-4) into the second equation:

-2A + (√(-4))B = 0.5

Simplifying, we get:

-2(0.25) - 2B√(-4) = 0.5
-0.5 - 4B√(-4) = 0.5
4B√(-4) = -1
B√(-4) = -1/4
B = (-1/4) / √(-4)
B = -1/4i

Now we can substitute the values of A and B back into the general solution:

x(t) = e^(-2t) * (0.25 * cos(ωt) + (-1/4i) * sin(ωt))

Simplifying, we get:

x(t) = 0.25 * e^(-2t) * cos(ωt) - (1/4) * e^(-2t) * i * sin(ωt)

Since ω = √(-4), we have:

ω = √4 * i = 2i

Finally, we can write:

x(t) = 0.25 * e^(-2t) * cos(2it) - (1/4) * e^(-2t) * i * sin(2it)

So, the position of the mass at time t is given by this equation.

To find the position of the mass at time t, we need to solve the differential equation for the motion of the mass.

The equation of motion for a damped harmonic oscillator is given by:

m * x''(t) + c * x'(t) + k * x(t) = 0

Where:
m = mass of the object (2 kg)
x(t) = position of the object at time t
x'(t) = velocity of the object at time t
x''(t) = acceleration of the object at time t
k = spring constant of the spring (to be determined)

We know that the force needed to keep the spring at a certain elongation is given by Hooke's Law:

F = k * x

Given that a force of 13 N is needed to keep the spring stretched by 0.25 m, we can solve for the spring constant k:

13 N = k * 0.25 m
k = 13 N / 0.25 m
k = 52 N/m

Now, let's rewrite the equation of motion using the known values:

2 kg * x''(t) + 8 * x'(t) + 52 * x(t) = 0

We also have the initial conditions for the motion of the mass:

x(0) = 0 (initial position)
x'(0) = 0.5 m/s (initial velocity)

To find the position of the mass at time t, we can solve this second-order ordinary differential equation using methods such as the Laplace transform, numerical integration, or a differential equation solver.

Using any method of solving differential equations, you can find the position x(t) as a function of time t.