Question

A mass weighing 12 pounds stretches a spring 2 feet. The massis initially released from a point 1 foot below the equilibriumposition with an upward velocity of 4 ft/s.

(a) Find the equation of motion.

(b) What are the amplitude, period, and frequency of thesimple harmonic motion?

(c) At what times does the mass return to the point 1 footbelow the equilibrium position?

(d) At what times does the mass pass through the equilibriumposition moving upward? Moving downward?

(e) What is the velocity of the mass at t 5 3y16 s?

(f) At what times is the velocity zero?

Short answer

(a) The equation of the motion is$x\left(t\right)=\cos 4t-\sin 4t$.

(b) The weight of the original mass on the spring is $14.4\text{lbs}$.

Step-by-step solution

Step 1:Determine the equation of the motion.

(a)

As there is no damping, then.$\beta =0$So, the differential equation of motion is defined asor $\frac{3}{8}\frac{d^{2}x}{d{t}^2}+6x=0$equivalently.$\frac{d^{2}x}{d{t}^2}+16x=0$

The auxiliary equation for the associated homogeneous equation is$m^2+16=0$. Solve the roots of the equation that gives $m_1=4i$and.$m_2=-4i$

Thus, the general solution of the differential equation is. $x\left(t\right)=c_1\cos 4t+c_2\sin 4t$Also, the first derivative of the above differential equation is.$x\text{'}\left(t\right)=-4c_1\sin 4t+4c_2\cos 4t$

As the mass is initially released from a point 1 foot below the equilibrium position with an upwards velocity of, the initial conditions will be $x\left(0\right)=1$and.$x\text{'}\left(0\right)=-4$Apply these conditions to get:

$\begin{array}{c}x\left(0\right)=c_1\cos 0+c_2\sin 0=1\\ c_1=1\end{array}$

$\begin{array}{c}x\text{'}\left(0\right)=-4c_1\sin 0+4c_2\cos 0=-4\\ 4c_2=-4\\ c_2=-1\end{array}$

Therefore, the equation of the motion is$x\left(t\right)=\cos 4t-\sin 4t$

Determine the amplitude, period, and frequency of the simple harmonic motion.

(b)

To find the amplitude, period, and frequency of the motion, convert a solution to the simpler form that is$x\left(t\right)=A\sin (\omega t+\varphi )$. Here, the amplitude is$A=\sqrt{c_1^2+c_2^2}$, and the phase angle is$\varphi ={\tan }^{-1}\frac{{\text{c}}_1}{c_2}$.

Hence, the amplitude and phase angle of the motion are:

$\begin{array}{c}A=\sqrt{1^2+{(-1)}^2}\\ =\sqrt{2}\\ \varphi ={\tan }^{-1}\frac{1}{-1}\\ =\frac{3\pi }{4}\end{array}$

Now, the equation of motion in simpler form is given by$x\left(t\right)=\sqrt{2}\sin (4t+\frac{3\pi }{4})$. Therefore, the period is:

$\begin{array}{c}T=\frac{2\pi }{\omega }\\ =\frac{2\pi }{4}\\ =\frac{\pi }{2}\text{s}\end{array}$

And the frequency is:

.$\begin{array}{c}f=\frac{1}{T}\\ =\frac{2}{\pi }\text{Hz}\end{array}$