Question

A mass weighing 39.5 kg. streches a spring 1/4m. At t=0 , the mass is released from a point 3/4m below the equilibrium position with an upward velocity of (5/4)m/sec. Determine the function x(t) that describes the subsiquent free motion.

Answer 1

Answer on Question #40031, Math, Differential Calculus

A mass weighing $39.5\,\mathrm{kg}$ stretches a spring $1/4\,\mathrm{m}$. At $t=0$, the mass is released from a point $3/4\,\mathrm{m}$ below the equilibrium position with an upward velocity of $(5/4)\,\mathrm{m/sec}$. Determine the function $x(t)$ that describes the subsequent free motion.

Solution.

$$m = 39.5\,\mathrm{kg}$$ $$L = \frac{1}{4}\,\mathrm{m}$$ $$x_0 = \frac{3}{4}\,\mathrm{m}$$ $$v_0 = x_0' = \frac{5}{4}\,\frac{\mathrm{m}}{\mathrm{s}}$$ $$g = 9.8\,\frac{\mathrm{m}}{\mathrm{s}^2}$$

For a mass $m$ suspended from a spring, the weight is jointly proportional to the distance stretched and a constant: $F = ks$ (Hooke's Law) or weight $= kx$.

If the motion of the mass is free and undamped (also called simple harmonic), it is described by:

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

Circular frequency: $\omega = \sqrt{\frac{k}{m}}$

To find the equation of motion:

$$\frac {d ^ {2} x}{d t ^ {2}} + \omega^ {2} x = 0$$

Let's $x = e^{\lambda t}$

Then

$$\lambda^ {2} e ^ {\lambda t} + \omega^ {2} e ^ {\lambda t} = 0$$ $$\lambda^ {2} + \omega^ {2} = 0$$ $$\lambda = \pm i \omega$$ $$x = C _ {1} e ^ {- i \omega t} + C _ {2} e ^ {i \omega t} = A \sin \omega t + B \cos \omega t$$ $$x (0) = x _ {0} = \frac {3}{4} - \text{initial displacement:}$$

positive if BELOW equlibrum point

negative if ABOVE equlibrum point

$$x ^ {\prime} (0) = x _ {0} ^ {\prime} = \frac {5}{4} - \text{initial velocity:}$$

positive DOWNWARD

negative UPWARD

$$= 0 \text{ if from rest}$$

So find $A$ and $B$:

$$x (0) = A \sin \omega \cdot 0 + B \cos \omega \cdot 0 = B = \frac {3}{4}$$ $$x ^ {\prime} (t) = A \omega \cos \omega t - B \omega \sin \omega t$$ $$x ^ {\prime} (0) = A \omega \cos \omega \cdot 0 - B \omega \sin \omega \cdot 0 = A \omega = \frac {5}{4}$$

So

$$A = \frac {5}{4 \omega}$$ $$B = \frac {3}{4}$$

Find $\omega$:

$$\omega = \sqrt {\frac {k}{m}}$$

From Hooke's Law:

$$F = k L = m g$$

where $L$ – point of equilibrium, $L = \frac{1}{4}$

$$k \cdot \frac{1}{4} m = 39.5 \, kg \cdot 9.8 \, \frac{m}{s^2}$$ $$k = 39.5 \cdot 9.8 \cdot 4 = 1548.4 \, \frac{N}{m} - \text{spring stiffness}$$ $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{L}} = 2\sqrt{g} \approx 6.3 \, \frac{\text{rad}}{s}$$

Therefore,

$$x = \frac{5}{4\omega} \sin \omega t + \frac{3}{4} \cos \omega t = 0.2 \sin 6.3 t + 0.75 \cos 6.3 t$$

Answer:

$$x(t) = 0.2 \sin 6.3 t + 0.75 \cos 6.3 t$$