It is the simplest case which we can consider. Unforced or free vibrations sense that F(t) = 0 and undamped vibrations implies that g = 0. Under this case the differential equation turn into,

mu′′ + ku = 0

It is easy sufficient to solve in common. The characteristic equation which has the roots,

r= + i √(k/m)

This is generally reduced to,

r= + w₀ i

Here,

w₀ = √(k/m)

And w₀ is termed as the natural frequency. Recall also that m > 0 and k > 0 and thus we can guarantee that such quantity will be complicated. The solution under this case is after that,

u(t)  = c₁ cos (w₀ t) + c₂ sin(w₀ t)     .........................................(4)

We can write (4) in the subsequent form,

u(t)  =R cos (w₀ t - d)   .........................(5)

Here R is the amplitude of the displacement and d is the phase shift or phase angle of the displacement.

While the displacement is in the form of (5) it is generally easier to work with. Though, it's easier to get the constants in (4) from the initial conditions so it is to determine the amplitude and phase shift in (5) from the initial conditions. Therefore, so as to get the equation in the form in (5) we will first put the equation into the form in (4), determine the constants, c₁ and c₂ and then convert it into the form in (5).

Therefore, assuming that we have c₁ and c₂ how do we find out R and d?  Let's begin with (5) and use a trig individuality to write this as

u(t)  =R cos (d) cos (w₀ t) + Rsin(d) sin(w₀ t)    ...............................(6)

Now, R and d are constants and therefore if we compare (6) to (4) we can notice that,

c₁ = R cos (d),                          c₂ = Rsin(d)

We can determine R in the following manner.

c²₁+c²₂ = R² cos²d + R² sin²d = R²

Both sides taking the square root of and assuming that R is positive will provide,

R = √(c²₁+c²₂)   ............................(7)

Finding d is just as simple.  We'll begin with,

c₂/c₁ = R sin(d)/ Rcos(d) = tan (d)

Both sides taking the inverse tangent gives,

d= tan^-1 (c₂/c₁)

Before we work any illustrations let's talk a little bit regarding to the units of mass and the British vs. metric system dissimilarities.

Recall this weight of the object is specified by

W = mg

Here m is the mass of the object and g is the gravitational acceleration. For the illustrations in this problem we'll be using the subsequent values for g.

 British: g = 32 ft / s²

 Metric: g = 9.8 m / s²

It is not the standard 32.2 ft/s² or 9.81 m/s², but by using these will create some of the numbers come out some nicer.

In the metric system the mass of objects is specified in kilograms (kg) and here is nothing for us to do. Though, in the British system we tend to be specified the weight of an object in pounds here keep in mind that pounds are the units of weight not mass and so we'll require to calculate the mass for these problems.