Question

The differential equation of a damped vibrating system under the action of an external periodic force is: d^2x/dt^2 +2 m0 dx/dt +n^2x = a cos pt Show that, if n>m0>0 the complementary function of the differential equation represents vibrations which are soon damped out. Find the particular integral in terms of periodic functions

Answer 1

Answer on Question #70334 – Math – Differential Equations

Question

The differential equation of a damped vibrating system under the action of an external periodic force is

$$\frac {d ^ {2} x}{d t ^ {2}} + 2 m _ {0} \frac {d x}{d t} + n ^ {2} x = a \cos p t$$

Show that, if $n > m_0 > 0$ the complementary function of the differential equation represents vibrations which are soon damped out. Find the particular integral in terms of periodic functions.

Solution

The characteristic equation is

$$\lambda^ {2} + 2 m _ {0} \lambda + n ^ {2} = 0,$$

Its roots are

$$\lambda = \frac {- 2 m _ {0} \pm \sqrt {4 m _ {0} ^ {2} - 4 n ^ {2}}}{2} = - m _ {0} \pm \sqrt {m _ {0} ^ {2} - n ^ {2}}$$

If $n > m_0 > 0$, then the characteristic equation has the complex roots, and the complementary function of the differential equation is

$$x = e ^ {- m _ {0} t} \left(c _ {1} \cos \left(t \sqrt {n ^ {2} - m _ {0} ^ {2}}\right) + c _ {2} \sin \left(t \sqrt {n ^ {2} - m _ {0} ^ {2}}\right)\right)$$

The multiplier $e^{-m_0t}$ means that vibrations are soon damped out since $m_0 > 0$.

The particular integral:

$$\tilde {x} = A \cos p t + B \sin p t$$

Then

$$\tilde {x} ^ {\prime} = - A p \sin p t + B p \cos p t$$ $$\tilde {x} ^ {\prime \prime} = - A p ^ {2} \cos p t - B p ^ {2} \sin p t$$ $$\tilde {x} ^ {\prime \prime} + 2 m _ {0} \tilde {x} ^ {\prime} + n ^ {2} \tilde {x} = - A p ^ {2} \cos p t - B p ^ {2} \sin p t + 2 m _ {0} (- A p \sin p t + B p \cos p t) + n ^ {2} (A \cos p t + B \sin p t) =$$ $$= a \cos p t$$ $$\left\{ \begin{array}{l} A n ^ {2} - A p ^ {2} - 2 m _ {0} B p = a \\ B n ^ {2} - B p ^ {2} - 2 m _ {0} A p = 0 \end{array} \right.$$ $$A = \frac {B}{2 m _ {0} p} (n ^ {2} - p ^ {2})$$ $$\frac {B}{2 m _ {0} p} (n ^ {2} - p ^ {2}) ^ {2} - 2 m _ {0} B p = a$$ $$B = \frac {2 a m _ {0} p}{(n ^ {2} - p ^ {2}) ^ {2} - 4 m _ {0} ^ {2} p ^ {2}}$$ $$A = \frac {a}{(n ^ {2} - p ^ {2}) ^ {2} - 4 m _ {0} ^ {2} p ^ {2}} (n ^ {2} - p ^ {2})$$

Answer:

$$\tilde {x} = \frac {a}{(n ^ {2} - p ^ {2}) ^ {2} - 4 m _ {0} ^ {2} p ^ {2}} (n ^ {2} - p ^ {2}) \cos p t + \frac {2 a m _ {0} p}{(n ^ {2} - p ^ {2}) ^ {2} - 4 m _ {0} ^ {2} p ^ {2}} \sin p t$$

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