# A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a...

## Question:

A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.

a. Express the equation of motion in the form {eq}\displaystyle x(t)=A\sin(\omega t + \phi). {/eq}

b. Express the equation of motion in the form {eq}\displaystyle x(t)=A\cos(\omega t - \phi). {/eq}

c. Use one of the solutions obtained in parts (a) and (b) to determine the times the mass attains a displacement below the equilibrium position numerically equal to 1/2 the amplitude of motion.

## Simple Harmonic Motion:

When reviewing spring and mass problems, the key points are to determine the variables that will determine if it is Simple Harmonic Motion or Non-Harmonic Motion. Because the movement is said to follow the equations provided (i.e. the distance from the equilibrium point is proportional to the force driving it), then it is an example of Simple Harmonic Motion.

## Answer and Explanation:

**Step 1:**

Determine the spring constant, k. Assume the spring moves only up and down (or in the y-direction) and relate the spring with spring constant, k, to Newton's Second Law to find the following equation: {eq}\begin{align*} \sum F_y &= mass \cdot acceleration \\ F_y &= m \cdot a \tag{1} \\ \end{align*} {/eq}

From Hooke's law, we know the relationship between the spring and its mass to be:

{eq}F = k \cdot x \tag{2} \\ {/eq} where x is the distance moved away from the equilibrium point. The information provided in the problem can be used with Hooke?s Law to determine the unknown spring constant, k. {eq}\begin{align*} F &= k \cdot x \tag{2} \\ 10 \ \rm{lbs} &= k \cdot 0.25 \ \rm{ft} \\ 40 \frac{\rm{lbs}}{\rm{ft}} &= k \\ \end{align*} {/eq}

**Step 2:**

Determine the equations of motion and the general solution. We know the relationship between position, x, and acceleration, a, to be: {eq}a = \frac{d^2x}{dt^2} \tag{3} \\ {/eq} Using equation (2), the spring constant, k, and equation (3) in equation (1) yields the following: {eq}\begin{align*} F_y &= m \cdot a \tag{1} \\ k \cdot x &= m \cdot \frac{d^2x}{dt^2} \\ 40 \frac{\rm{lbs}}{\rm{ft}} \cdot x &= m \cdot \frac{d^2x}{dt^2} \\ 0 &= m \cdot \frac{d^2x}{dt^2}+40 \frac{\rm{lbs}}{\rm{ft}} \cdot x \\ 0 &= \frac{d^2x}{dt^2}+ \frac{40 \frac{\rm{lbs}}{\rm{ft}}}{m} \cdot x \tag{4} \\ \end{align*} {/eq}

Because the problem requires information on the movement of the spring with {eq}m_2 = 1.6 \ \rm{slugs} {/eq}, we can use this in the differential equation. Note: Although the units of the two masses provided are different, they do not interfere in the problem because their units cancel out nicely. {eq}\begin{align*} 0 &= \frac{d^2x}{dt^2}+ \frac{40 \frac{\rm{lbs}}{\rm{ft}}}{m} \cdot x \tag{4} \\ 0 &= \frac{d^2x}{dt^2}+ \frac{40 \frac{\rm{lbs}}{\rm{ft}}}{1.6 \ \rm{slugs}} \cdot x \\ 0 &= \frac{d^2x}{dt^2}+ 25 \cdot x \tag{5} \\ \end{align*} {/eq}

This differential equation looks similar to {eq}0 = m \cdot \frac{d^2x}{dt^2}+\omega^2 \cdot x \\ {/eq} which has a general solution of {eq}x(t)=c_1 \cdot cos(\omega t)+c_2 \cdot sin(\omega t) \\ {/eq} where {eq}\omega^2 = 25 {/eq}. The general solution is re-written to be: {eq}\begin{align*} x(t) &=c_1 \cdot cos(5 \cdot t)+c_2 \cdot sin(5 \cdot t) \tag{6} \\ \frac{dx}{dt} &= -5 \cdot c_1 \cdot sin(5 \cdot t)+5 \cdot c_2 \cdot cos(5 \cdot t) \tag{7} \\ \end{align*} {/eq}

Using the initial conditions provided:

{eq}\begin{align*} x(0) &= -\frac{1}{3} \ \rm{ft} \\ v(0) &= \frac{dx}{dt} = \frac{5}{4} \\ \end{align*} {/eq}

The equations of motion are found to be the following: {eq}\begin{align*} x(t) &=c_1 \cdot cos(5 \cdot t)+c_2 \cdot sin(5 \cdot t) \tag{6} \\ -\frac{1}{3} \ \rm{ft} &=c_1 \cdot cos(5 \cdot 0)+c_2 \cdot sin(5 \cdot 0) \\ -\frac{1}{3} \ \rm{ft} &=c_1 \cdot 1+c_2 \cdot 0 \\ -\frac{1}{3} &=c_1 \\ \\ \frac{dx}{dt} &= -5 \cdot c_1 \cdot sin(5 \cdot t)+5 \cdot c_2 \cdot cos(5 \cdot t) \tag{7} \\ \frac{5}{4} &= -5 \cdot - \frac{1}{3} \cdot sin(5 \cdot 0)+5 \cdot c_2 \cdot cos(5 \cdot 0) \\ \frac{5}{4} &= 0+5 \cdot c_2 \cdot 1 \\ \frac{1}{4} &= c_2 \\ \end{align*} {/eq}

Substituting these values into the general solution (6), we will get: {eq}x(t) =-\frac{1}{3} \cdot cos(5 \cdot t)+\frac{1}{4} \cdot sin(5 \cdot t) \tag{8} \\ {/eq}

**Step 3:**

Solve for part a. To express the equation of motion in the {eq}\displaystyle x(t)=A\sin(\omega t + \phi) {/eq} format, we must find the amplitude, {eq}A {/eq}, and the phase angle, {eq}\phi {/eq}. The amplitude is defined as:

{eq}\begin{align*} A &= \sqrt{c_1^2+c_2^2} \\ A &= \sqrt{(-\frac{1}{3})^2+(\frac{1}{4})^2} \\ A &= \frac{5}{12} \\ \end{align*} {/eq} Knowing {eq}sin \phi = \frac{c_1}{A} < 0 {/eq} and {eq}cos \phi = \frac{c_2}{A} > 0 {/eq} and {eq}tan \phi \frac{c_1}{c_2} = -\frac{4}{3} {/eq}, we know this solution lies in the fourth quadrant. The phase angle {eq}\phi {/eq} is found:

{eq}\begin{align*} \phi &= tan_{-1} \frac{c_1}{c_2} \\ \phi &= tan_{-1} \frac{-\frac{1}{3}}{\frac{1}{4}} \\ \phi &= -0.927 \\ \end{align*} {/eq}

Therefore, the answer to part a will be: {eq}\displaystyle x(t)=\frac{5}{12} \sin(5t-0.927) \\ {/eq}

**Step 4.**

Solve for part b. To express the equation of motion in the {eq}\displaystyle x(t)=A\cos(\omega t - \phi). {/eq} format, we must find the new phase angle, {eq}\phi {/eq}. The amplitude remains unchanged. Knowing {eq}sin \phi = \frac{c_2}{A} > 0 {/eq} and {eq}cos \phi = \frac{c_1}{A} < 0 {/eq} and {eq}tan \phi \frac{c_2}{c_1} = -\frac{3}{4} {/eq}, we know this solution lies in the second quadrant. The phase angle can now be determined:

The phase angle {eq}\phi {/eq} is found:

{eq}\begin{align*} \phi &= tan_{-1} \frac{c_2}{c_1} + \phi \\ \phi &= tan_{-1} \frac{\frac{1}{4}}{-\frac{1}{3}} + \phi\\ \phi &= -0.6434 + \phi \\ \phi &= 2.498 \\ \end{align*} {/eq} Therefore, the answer to part b will be: {eq}\displaystyle x(t)=\frac{5}{12} \cos(5t-2.498) \\ {/eq}

**Step 5:**

Solve for part c. At {eq}x(t) = \frac{1}{2} \cdot A {/eq} we can set x(t) equal to either equation found previously in (a) or (b). When set to the equation found in part (a), the solution is obtained as shown below:

{eq}\begin{align*} x(t) &= A \sin(5t-0.927) \\ \frac{1}{2} \cdot A &= A \sin(5t-0.927) \\ \frac{\pi}{6}+2k\pi, \frac{5\pi}{6}+2n\pi &= 5t-0.927 \\ 0.290 \cdot \frac{\pi}{6}+2k\pi, 0.709 \cdot \frac{5\pi}{6}+2n\pi &= t \\ \end{align*} {/eq}

where {eq}n = 0, 1, 2, 3, ... {/eq}.

So, {eq}t = 0.290 \ \rm{s}, 1.547 \ \rm{s}, 2.803 \ \rm{s} ... {/eq} and {eq}t = 0.709 \ \rm{s}, 1.966 \ \rm{s}, 3.222 \ \rm{s}... {/eq}.

Therefore: {eq}\boxed {a. x(t)=\frac{5}{12} \sin(5t-0.927)} \\ \boxed {b. x(t)=\frac{5}{12} \cos(5t-2.498)} \\ \boxed {c. t = 0.290 \ \rm{s}, 1.547 \ \rm{s}, 2.803 \ \rm{s}, ... and \ t = 0.709 \ \rm{s}, 1.966 \ \rm{s}, 3.222 \ \rm{s}, ...} {/eq}

#### Learn more about this topic:

from AP Physics 1: Exam Prep

Chapter 8 / Lesson 2