Question

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

Answer 1

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

$A = 1.803 ft$

Period = $\frac{\pi}{2}$ seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = $\frac{w}{g}$

m= $\frac{32}{32}$

= 1 slug

Also from Hooke's Law

2 k = 32

k = $\frac{32}{2}$

k = 16 lb/ft

Using the function:

$\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0$

$x(0) = -1$        (because of the initial position being above the equilibrium position)

$x(0) = -6$          ( as a result of upward velocity)

NOW, we have:

$x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)$

However;

$x(0) = -1$ means

$-1 =c_1\\c_1 = -1$

$x(0) =-6$ also implies that:

$-6 =4(c_2)\\c_2 = - \frac{6}{4}$

$c_2 = -\frac{3}{2}$

Hence, $x(t) =-cos4t-\frac{3}{2} sin 4t$

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

$A = \sqrt{(-1)^2+(\frac{3}{2})^2 }$

$A=\sqrt{\frac{13}{4} }$

$A= \frac{1}{2}\sqrt{13}$

$A = 1.803 ft$

Period can be calculated as follows:

= $\frac{2 \pi}{4}$

= $\frac{\pi}{2}$ seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

$x* \frac{\pi}{2} = 4 \pi$

$x \pi = 8 \pi$

$x=8$ cycles