Question

A spring with a 4 kg mass is resting on a horizontal frictionless table and is attached to a wall at the left end. It will stretch 0.5 m beyond its natural length when a force of 12 N is applied. If the spring begins at the equilibrium position and is pushed to the right with an initial velocity of 3 m/s, find the position of the mass after t seconds.

Answer 1

Answer:

The position of the mass after t seconds is;

$x(t) = 1.225 sin(2.45t)$

Explanation:

Let x(t) represent the position after t seconds

Apply the following differential equation;

$m\frac{d^2x}{dt^2} + kx = 0$

where;

m is the mass of the spring = 4 kg

k is the spring constant, = F/L = 12 / 0.5 = 24 N/m

the, ω² = k/m = 24 / 4 = 6

ω = √6 = 2.45

The general solution is given by;

$x(t) = C_1 cos(\omega t) + C_2 sin (\omega t)$

$x(t) = C_1 cos(2.45 t) + C_2 sin (2.45 t)\\\\then, x'(t) = -2.45C_1sin(2.45 t) + 2.45C_2cos(2.45t)$

at x(0) = 0 , C₁ = 0    and

at x'(0) = 3 m/s , C₂ = 1.225

Therefore, the position of the mass after t seconds is given by;

$x(t) = 1.225 sin(2.45t)$

Answer 2

The position of the given spring mass after t seconds is;

x(t) = 1.22 sin 2.45t

We are given;

Mass of spring; m = 4 kg

Force; F = 12 N

Extension of spring; x = 0.5 m

Now, the formula for the force in the spring is;

F = kx

Where k is spring constant

Thus; k = F/x

k = 12/0.5

k = 24 N/m

Formula for the angular velocity is;

ω = k/m

ω² = 24/4

ω = √6

ω = 2.45

Thus, the general solution is;

x(t) = C₁ cos ωt + C₂ sin ωt

⇒ x(t) = C₁ cos 2.45t + C₂ sin 2.45t

Applying initial condition of x(0) = 0 gives;

C₁ = 0

Let's now find the derivative of x(t) which is the velocity;

x'(t) = -2.45C₁ sin 2.45t  + 2.45C₂ cos 2.45t

Applying the initial condition x'(0) = 3 gives;

2.45C₂ = 3

C₂ = 3/2.45

C₂ = 1.22

Thus, in conclusion the the position of the mass after t seconds is;

x(t) = 1.22 sin 2.45t

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