Question

The differential equation of a spring/mass system is $x^n+16x=0$. If the mass is initially released from a point 1 meter above the equilibrium position with a downward velocity of 3 m/s, the amplitude of vibrations is _________meters.

Short answer

The given equation of a spring/mass system is$x^n+16x=0$ .

If the mass is initially released from a point 1 meter above the equilibrium position with a downward velocity of 3 m/s, the amplitude of vibrations is$\frac{5}{4}$ meters.

Step-by-step solution

Definition of spring / mass system

The period of whatever object moving in a simple harmonic motion can generally be determined using the spring-mass system.

The starting criteria are as follows:

The mass is discharged from a location one metre above the equilibrium position, thus $x\left(0\right)=-1$.

Because the mass is expelled at a 3 m/s downward velocity, $x\text{'}\left(0\right)=3$.

Step 2:

Let the given equation as equation (1)

$x^n+16x=0············\left(1\right)$

The general solution of equation (1) is below

$x\left(t\right)=c_1\cos 4t+c_2\sin 4t··············\left(2\right)$

As a starting criterion gives,

$c_1=-1$

So, the equation (2) is becoming

$x\left(t\right)=-\cos 4t+c_2\sin 4t··············\left(3\right)$

Step 3:

The second criteria give us

$c_2=\frac{3}{4}$

Now, apply the value to equation (3)

$x\left(t\right)=-\cos 4t+\frac{3}{4}\sin 4t··············\left(3\right)$

Find the amplitude and given by

$A=\sqrt{C_1^2+C_2^2}$

So,

$$\begin{gathered} A=\sqrt{1+\frac{9}{16}} \\ A=\frac{5}{4} m \end{gathered}$$

Therefore, the amplitude of vibrations is $\frac{5}{4}$meters.