6.2 Spring Problems II

Free Vibrations With Damping

In this section we consider the motion of an object in a spring–mass system with damping. We start with unforced motion, so the equation of motion is

\begin{equation} my''+cy'+ky=0. \tag{6.2.1}\end{equation}

Now suppose the object is displaced from equilibrium and given an initial velocity. Intuition suggests that if the damping force is sufficiently weak the resulting motion will be oscillatory, as in the undamped case considered in the previous section, while if it’s sufficiently strong the object may just move slowly toward the equilibrium position without ever reaching it. We’ll now confirm these intuitive ideas mathematically. The characteristic equation of (6.2.1) is

\[ mr^2+cr+k=0. \]

The roots of this equation are

\begin{equation} r_1={-c-\sqrt{c^2-4mk}\over2m}\mbox{\quad and \quad} r_2= {-c+\sqrt{c^2-4mk}\over2m}. \tag{6.2.2}\end{equation}

In Section 5.2 we saw that the form of the solution of (6.2.1) depends upon whether \(c^2-4mk\) is positive, negative, or zero. We’ll now consider these three cases.

Underdamped Motion

We say the motion is underdamped if \(c<\sqrt{4mk}\). In this case \(r_1\) and \(r_2\) in (6.2.2) are complex conjugates, which we write as

\[ r_1=-{c\over2m}-i\omega_1\mbox{\quad and \quad} r_2=-{c\over2m}+i\omega_1, \]

where

\[ \omega_1={\sqrt{4mk-c^2}\over2m}. \]

The general solution of (6.2.1) in this case is

\[ y=e^{-ct/2m}(c_1\cos\omega_1 t+c_2\sin\omega_1 t). \]

By the method used in Section 6.1 to derive the amplitude–phase form of the displacement of an object in simple harmonic motion, we can rewrite this equation as

\begin{equation} y=Re^{-ct/2m}\cos(\omega_1 t-\phi), \tag{6.2.3}\end{equation}

where

\[ R=\sqrt{c_1^2+c_2^2},\quad R\cos\phi=c_1,\mbox{\quad and \quad} R\sin\phi=c_2. \]

The factor \(Re^{-ct/2m}\) in (6.2.3) is called the time–varying amplitude of the motion, the quantity \(\omega_1\) is called the frequency, and \(T=2\pi/\omega_1\) (which is the period of the cosine function in (6.2.3) is called the quasi–period. A typical graph of (6.2.3) is shown in Figure 6.2.1. As illustrated in that figure, the graph of \(y\) oscillates between the dashed exponential curves \(y=\pm Re^{-ct/2m}\).

Overdamped Motion

We say the motion is overdamped if \(c>\sqrt{4mk}\). In this case the zeros \(r_1\) and \(r_2\) of the characteristic polynomial are real, with \(r_1<r_2<0\) (see (6.2.2)), and the general solution of (6.2.1) is

\[ y=c_1e^{r_1t}+c_2e^{r_2t}. \]

Again \(\lim_{t\to\infty}y(t)=0\) as in the underdamped case, but the motion isn’t oscillatory, since \(y\) can’t equal zero for more than one value of \(t\) unless \(c_1=c_2=0\). (Exercise 23.)

Critically Damped Motion

We say the motion is critically damped if \(c=\sqrt{4mk}\). In this case \(r_1=r_2=-c/2m\) and the general solution of (6.2.1) is

\[ y=e^{-ct/2m}(c_1+c_2t). \]

Again \(\lim_{t\to\infty}y(t)=0\) and the motion is nonoscillatory, since \(y\) can’t equal zero for more than one value of \(t\) unless \(c_1=c_2=0\). (Exercise 22).

Example 6.2.1

Suppose a 64 lb weight stretches a spring 6 inches in equilibrium and a dashpot provides a damping force of \(c\) lb for each ft/sec of velocity.

Write the equation of motion of the object and determine the value of \(c\) for which the motion is critically damped.
Find the displacement \(y\) for \(t>0\) if the motion is critically damped and the initial conditions are \(y(0)=1\) and \(y'(0)=20\).
Find the displacement \(y\) for \(t>0\) if the motion is critically damped and the initial conditions are \(y(0)=1\) and \(y'(0)=-20\).

Solution (a) Here \(m=2\) slugs and \(k=64/.5=128\) lb/ft. Therefore the equation of motion (6.2.1) is

\begin{equation} 2y''+cy'+128y=0. \tag{6.2.4}\end{equation}

The characteristic equation is

\[ 2r^2+cr+128=0, \]

which has roots

\[ r={-c\pm\sqrt{c^2-8\cdot128}\over4}. \]

Therefore the damping is critical if

\[ c=\sqrt{8\cdot128}=32\mbox{ lb--sec/ft}. \]

Solution (b) Setting \(c=32\) in (6.2.4) and cancelling the common factor \(2\) yields

\[ y''+16y+64y=0. \]

The characteristic equation is

\[ r^2+16r+64y=(r+8)^2=0. \]

Hence, the general solution is

\begin{equation} y=e^{-8t}(c_1+c_2t). \tag{6.2.5}\end{equation}

Differentiating this yields

\begin{equation} y'=-8y+c_2e^{-8t}. \tag{6.2.6}\end{equation}

Imposing the initial conditions \(y(0)=1\) and \(y'(0)=20\) in the last two equations shows that \(1=c_1\) and \(20=-8+c_2\). Hence, the solution of the initial value problem is

\[ y=e^{-8t}(1+28t). \]

Therefore the object approaches equilibrium from above as \(t\to\infty\). There’s no oscillation. Solution (c) Imposing the initial conditions \(y(0)=1\) and \(y'(0)=-20\) in (6.2.5) and (6.2.6) yields \(1=c_1\) and \(-20=-8+c_2\). Hence, the solution of this initial value problem is

\[ y=e^{-8t}(1-12t). \]

Therefore the object moves downward through equilibrium just once, and then approaches equilibrium from below as \(t\to\infty\). Again, there’s no oscillation. The solutions of these two initial value problems are graphed in Figure 6.2.2.

Example 6.2.2

Find the displacement of the object in Example 6.2.1 if the damping constant is \(c=4\) lb–sec/ft and the initial conditions are \(y(0)=1.5\) ft and \(y'(0)=-3\) ft/sec.

Solution With \(c=4\), the equation of motion (6.2.4) becomes

\begin{equation} y''+2y'+64y=0 \tag{6.2.7}\end{equation}

after cancelling the common factor 2. The characteristic equation

\[ r^2+2r+64=0 \]

has complex conjugate roots

\[ r={-2\pm\sqrt{4-4\cdot64}\over2}=-1\pm3\sqrt7i. \]

Therefore the motion is underdamped and the general solution of (6.2.7) is

\[ y=e^{-t}(c_1\cos3\sqrt7t+c_2\sin3\sqrt7t). \]

Differentiating this yields

\[ y'=-y+3\sqrt7e^{-t}(-c_1\sin3\sqrt7t+c_2\cos3\sqrt7t). \]

Imposing the initial conditions \(y(0)=1.5\) and \(y'(0)=-3\) in the last two equations yields \(1.5=c_1\) and \(-3=-1.5+3\sqrt7c_2\). Hence, the solution of the initial value problem is

\begin{equation} y=e^{-t}\bigg({3\over2}\cos3\sqrt7t-{1\over2\sqrt7} \sin3\sqrt7t\bigg). \tag{6.2.8}\end{equation}

The amplitude of the function in parentheses is

\[ R=\sqrt{\bigg({3\over2}\bigg)^2+\bigg({1\over2\sqrt7}\bigg)^2} =\sqrt{{9\over4}+{1\over4\cdot7}} =\sqrt{64\over4\cdot7}={4\over\sqrt7}. \]

Therefore we can rewrite (6.2.8) as

\[ y={4\over\sqrt7}e^{-t}\cos(3\sqrt7t-\phi), \]

where

\[ \cos\phi={3\over2R}={3\sqrt7\over8}\mbox{\quad and \quad}\sin\phi=-{1\over2\sqrt7R}= -{1\over8}. \]

Therefore \(\phi\cong-.125\) radians.

Example 6.2.3

Let the damping constant in Example 1 be \(c=40\) lb–sec/ft. Find the displacement \(y\) for \(t>0\) if \(y(0)=1\) and \(y'(0)=1\).

Solution With \(c=40\), the equation of motion (6.2.4) reduces to

\begin{equation} y''+20y'+64y=0 \tag{6.2.9}\end{equation}

after cancelling the common factor 2. The characteristic equation

\[ r^2+20r+64=(r+16)(r+4)=0 \]

has the roots \(r_1=-4\) and \(r_2=-16\). Therefore the general solution of (6.2.9) is

\begin{equation} y=c_1e^{-4t}+c_2e^{-16t}. \tag{6.2.10}\end{equation}

Differentiating this yields

\[ y'=-4e^{-4t}-16c_2e^{-16t}. \]

The last two equations and the initial conditions \(y(0)=1\) and \(y'(0)=1\) imply that

\[ \begin{array}{rlrl} c_1&+&c_2&=1\\ -4c_1&-&16c_2&=1. \end{array} \]

The solution of this system is \(c_1=17/12\), \(c_2=-5/12\). Substituting these into (6.2.10) yields

\[ y={17\over12}e^{-4t}-{5\over12}e^{-16t} \]

as the solution of the given initial value problem (Figure 6.2.3).

y=1712e^-4t-512e^-16t — Figure 6.2.3. \(y=\dst{{17\over12}e^{-4t}-{5\over12}e^{-16t}}\)

Forced Vibrations With Damping

Now we consider the motion of an object in a spring-mass system with damping, under the influence of a periodic forcing function \(F(t)=F_0\cos\omega t\), so that the equation of motion is

\begin{equation} my''+cy'+ky=F_0\cos\omega t. \tag{6.2.11}\end{equation}

In Section 6.1 we considered this equation with \(c=0\) and found that the resulting displacement \(y\) assumed arbitrarily large values in the case of resonance (that is, when \(\omega=\omega_0=\sqrt{k/m}\)). Here we’ll see that in the presence of damping the displacement remains bounded for all \(t\), and the initial conditions have little effect on the motion as \(t\to\infty\). In fact, we’ll see that for large \(t\) the displacement is closely approximated by a function of the form

\begin{equation} y=R\cos(\omega t-\phi), \tag{6.2.12}\end{equation}

where the amplitude \(R\) depends upon \(m\), \(c\), \(k\), \(F_0\), and \(\omega\). We’re interested in the following question:

Question: Assuming that \(m\), \(c\), \(k\), and \(F_0\) are held constant, what value of \(\omega\) produces the largest amplitude \(R\) in (6.2.12), and what is this largest amplitude?

To answer this question, we must solve (6.2.11) and determine \(R\) in terms of \(F_0,\omega_0,\omega\), and \(c\). We can obtain a particular solution of (6.2.11) by the method of undetermined coefficients. Since \(\cos\omega t\) does not satisfy the complementary equation

\[ my''+cy'+ky=0, \]

we can obtain a particular solution of (6.2.11) in the form

\begin{equation} y_p=A\cos\omega t+B\sin\omega t. \tag{6.2.13}\end{equation}

Differentiating this yields

\[ y_p'=\quad\omega (-A\sin\omega t+B\cos\omega t) \]

and

\[ y_p''=-\omega^2(A\cos\omega t+B\sin\omega t). \]

From the last three equations,

\[ my''_p+cy'_p+ky_p=(-m\omega^2A+c\omega B+kA)\cos\omega t+ (-m\omega^2 B-c\omega A+kB)\sin\omega t, \]

so \(y_p\) satisfies (6.2.11) if

\[ \begin{array}{lll} (k-m\omega^2) A+\quad c\omega B &=&F_0\\ -c\omega A\quad+(k-m\omega^2)B&=&0. \end{array} \]

Solving for \(A\) and \(B\) and substituting the results into (6.2.13) yields

\[ y_p={F_0\over(k-m\omega^2)^2+c^2\omega^2} \left[(k-m\omega^2)\cos\omega t+c\omega\sin\omega t\right], \]

which can be written in amplitude–phase form as

\begin{equation} y_p={F_0\over\sqrt{(k-m\omega^2)^2+c^2\omega^2}} \cos(\omega t-\phi), \tag{6.2.14}\end{equation}

where

\begin{equation} \cos\phi={k-m\omega^2\over\sqrt {(k-m\omega^2)^2+c^2\omega^2}}\mbox{\quad and \quad} \sin\phi={c\omega\over\sqrt{(k-m\omega^2)^2+c^2\omega^2}}. \tag{6.2.15}\end{equation}

To compare this with the undamped forced vibration that we considered in Section 6.1 it’s useful to write

\begin{equation} k-m\omega^2=m\bigg({k\over m}-\omega^2\bigg)= m(\omega_0^2-\omega^2), \tag{6.2.16}\end{equation}

where \(\omega_0=\sqrt{k/m}\) is the natural angular frequency of the undamped simple harmonic motion of an object with mass \(m\) on a spring with constant \(k\). Substituting (6.2.16) into (6.2.14) yields

\begin{equation} y_p={F_0\over\sqrt{m^2(\omega^2_0-\omega^2)^2+ c^2\omega^2}}\cos(\omega t-\phi). \tag{6.2.17}\end{equation}

The solution of an initial value problem

\[ my''+cy'+ky=F_0\cos\omega t, \quad y(0)=y_0,\quad y'(0)=v_0, \]

is of the form \(y=y_c+y_p\), where \(y_c\) has one of the three forms

\begin{eqnarray*} y_c&=&e^{-ct/2m}(c_1\cos\omega_1t+c_2\sin\omega_1t), \\ y_c&=&e^{-ct/2m}(c_1+c_2t), \\ y_c&=&c_1e^{r_1t}+c_2e^{r_2t}\,(r_1,r_2<0). \end{eqnarray*}

In all three cases \(\lim_{t\to\infty} y_c(t)=0\) for any choice of \(c_1\) and \(c_2\). For this reason we say that \(y_c\) is the transient component of the solution \(y\). The behavior of \(y\) for large \(t\) is determined by \(y_p\), which we call the steady state component of \(y\). Thus, for large \(t\) the motion is like simple harmonic motion at the frequency of the external force.

The amplitude of \(y_p\) in (6.2.17) is

\begin{equation} R={F_0\over\sqrt{m^2(\omega^2_0-\omega^2)^2+c^2\omega^2}}, \tag{6.2.18}\end{equation}

which is finite for all \(\omega\); that is, the presence of damping precludes the phenomenon of resonance that we encountered in studying undamped vibrations under a periodic forcing function. We’ll now find the value \(\omega_{\max}\) of \(\omega\) for which \(R\) is maximized. This is the value of \(\omega\) for which the function

\[ \rho (\omega)=m^2(\omega^2_0-\omega^2)^2+c^2\omega^2 \]

in the denominator of (6.2.18) attains its minimum value. By rewriting this as

\begin{equation} \rho (\omega)=m^2(\omega^4_0+\omega^4)+ (c^2-2m^2\omega^2_0)\omega^2, \tag{6.2.19}\end{equation}

you can see that \(\rho\) is a strictly increasing function of \(\omega^2\) if

\[ c\ge\sqrt{2m^2\omega^2_0}=\sqrt{2mk}. \]

(Recall that \(\omega^2_0=k/m\)). Therefore \(\omega_{\max}=0\) if this inequality holds. From (6.2.15), you can see that \(\phi=0\) if \(\omega=0\). In this case, (6.2.14) reduces to

\[ y_p={F_0\over\sqrt{m^2\omega^4_0}}={F_0\over k}, \]

which is consistent with Hooke’s law: if the mass is subjected to a constant force \(F_0\), its displacement should approach a constant \(y_p\) such that \(ky_p=F_0\). Now suppose \(c<\sqrt{2mk}\). Then, from (6.2.19),

\[ \rho'(\omega)=2\omega(2m^2\omega^2+c^2-2m^2\omega^2_0), \]

and \(\omega_{\max}\) is the value of \(\omega\) for which the expression in parentheses equals zero; that is,

\[ \omega_{\max}=\sqrt{\omega^2_0-{c^2\over2m^2}} =\sqrt{{k\over m}\left(1-{c^2\over2km}\right)}. \]

(To see that \(\rho(\omega_{\max})\) is the minimum value of \(\rho(\omega)\), note that \(\rho'(\omega)<0\) if \(\omega <\omega_{\max}\) and \(\rho'(\omega)>0\) if \(\omega>\omega_{\max}\).) Substituting \(\omega=\omega_{\max}\) in (6.2.18) and simplifying shows that the maximum amplitude \(R_{\max}\) is

\[ R_{\max}={2mF_0\over c\sqrt{4mk-c^2}} \mbox{\quad if \quad} c< \sqrt{2mk}. \]

We summarize our results as follows.

Theorem 6.2.1

Suppose we consider the amplitude \(R\) of the steady state component of the solution of

\[ my''+cy'+ky=F_0\cos\omega t \]

as a function of \(\omega\). (a) If \(c\ge\sqrt{2mk}\), the maximum amplitude is \(R_{\max}=F_0/k\) and it’s attained when \(\omega= \omega_{\max}=0\). (b) If \(c<\sqrt{2mk}\), the maximum amplitude is

\begin{equation} R_{\max}={2m F_0\over c\sqrt{4mk-c^2}}, \tag{6.2.20}\end{equation}

and it’s attained when

\begin{equation} \omega=\omega_{\max}=\sqrt{{k\over m}\left(1-{c^2\over 2km}\right)}. \tag{6.2.21}\end{equation}

Note that \(R_{\max}\) and \(\omega_{\max}\) are continuous functions of \(c\), for \(c\ge0\), since (6.2.20) and (6.2.21) reduce to \(R_{\max}=F_0/k\) and \(\omega_{\max}=0\) if \(c=\sqrt{2km}\).

6.2 Exercises

A 64 lb object stretches a spring 4 ft in equilibrium. It is attached to a dashpot with damping constant \(c=8\) lb-sec/ft. The object is initially displaced 18 inches above equilibrium and given a downward velocity of 4 ft/sec. Find its displacement and time–varying amplitude for \(t>0\).

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\(y=\dst{{e^{-2t}\over2}(3\cos2t-\sin2t)\mbox{ ft}; \; \sqrt{{5\over2}} e^{-2t}}\) ft
C/G A 16 lb weight is attached to a spring with natural length 5 ft. With the weight attached, the spring measures 8.2 ft. The weight is initially displaced 3 ft below equilibrium and given an upward velocity of 2 ft/sec. Find and graph its displacement for \(t>0\) if the medium resists the motion with a force of one lb for each ft/sec of velocity. Also, find its time–varying amplitude.

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\(y=\dst{-e^{-t}\left(3\cos3t+{1\over3}\sin 3t\right)\mbox{ ft}\; {\sqrt{82}\over3}e^{-t}}\) ft
C/G An 8 lb weight stretches a spring 1.5 inches. It is attached to a dashpot with damping constant \(c\)=8 lb-sec/ft. The weight is initially displaced 3 inches above equilibrium and given an upward velocity of 6 ft/sec. Find and graph its displacement for \(t>0\).

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\(y=\dst{e^{-16t}\left({1\over4}+10t\right)}\) ft
A 96 lb weight stretches a spring 3.2 ft in equilibrium. It is attached to a dashpot with damping constant \(c\)=18 lb-sec/ft. The weight is initially displaced 15 inches below equilibrium and given a downward velocity of 12 ft/sec. Find its displacement for \(t>0\).

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\(y=\dst{-{e^{-3t}\over4}(5\cos t+63\sin t)}\) ft
A 16 lb weight stretches a spring 6 inches in equilibrium. It is attached to a damping mechanism with constant \(c\). Find all values of \(c\) such that the free vibration of the weight has infinitely many oscillations.

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\(0\le c<8\) lb-sec/ft
An 8 lb weight stretches a spring .32 ft. The weight is initially displaced 6 inches above equilibrium and given an upward velocity of 4 ft/sec. Find its displacement for \(t>0\) if the medium exerts a damping force of 1.5 lb for each ft/sec of velocity.

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\(y=\dst{{1\over2}e^{-3t}\left(\cos\sqrt{91}t+{11\over \sqrt{91}}\sin\sqrt{91}t\right)}\) ft
A 32 lb weight stretches a spring 2 ft in equilibrium. It is attached to a dashpot with constant \(c=8\) lb-sec/ft. The weight is initially displaced 8 inches below equilibrium and released from rest. Find its displacement for \(t>0\).

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\(y=-\dst{{e^{-4t}\over3} (2+8t)}\) ft
A mass of 20 gm stretches a spring 5 cm. The spring is attached to a dashpot with damping constant 400 dyne sec/cm. Determine the displacement for \(t>0\) if the mass is initially displaced 9 cm above equilibrium and released from rest.

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\(y=\dst{e^{-10t}\left(9\cos4\sqrt{6}t+{45\over 2\sqrt{6}}\sin4\sqrt{6} t\right)}\) cm
A 64 lb weight is suspended from a spring with constant \(k=25\) lb/ft. It is initially displaced 18 inches above equilibrium and released from rest. Find its displacement for \(t>0\) if the medium resists the motion with 6 lb of force for each ft/sec of velocity.

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\(\dst{y=e^{-3t/2}\left({3\over2}\cos {\sqrt{41}\over2}t +{9\over2\sqrt{41}}\sin{\sqrt{41}\over2}t\right)}\) ft
A 32 lb weight stretches a spring 1 ft in equilibrium. The weight is initially displaced 6 inches above equilibrium and given a downward velocity of 3 ft/sec. Find its displacement for \(t>0\) if the medium resists the motion with a force equal to 3 times the speed in ft/sec.

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\(y=\dst{e^{-{3\over2}t}\left({1\over2}\cos {\sqrt{119}\over2} t-{9\over2\sqrt{119}}\sin{\sqrt{119}\over2}t\right)}\) ft
An 8 lb weight stretches a spring 2 inches. It is attached to a dashpot with damping constant \(c\)=4 lb-sec/ft. The weight is initially displaced 3 inches above equilibrium and given a downward velocity of 4 ft/sec. Find its displacement for \(t>0\).

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\(\dst{y=e^{-8t}\left({1\over4}\cos8\sqrt{2}t-{1 \over4\sqrt{2}}\sin8\sqrt{2}t\right)}\) ft
C/G A 2 lb weight stretches a spring .32 ft. The weight is initially displaced 4 inches below equilibrium and given an upward velocity of 5 ft/sec. The medium provides damping with constant \(c=1/8\) lb-sec/ft. Find and graph the displacement for \(t>0\).

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\(\dst{y=e^{-t}\left(-{1\over3}\cos3\sqrt{11} t+ {14\over9\sqrt{11}}\sin3\sqrt{11}t\right)}\) ft
An 8 lb weight stretches a spring 8 inches in equilibrium. It is attached to a dashpot with damping constant \(c=.5\) lb-sec/ft and subjected to an external force \(F(t)=4\cos2t\) lb. Determine the steady state component of the displacement for \(t>0\).

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\(\dst{y_p={22\over61}\cos2t+{2\over61}\sin2t}\) ft
A 32 lb weight stretches a spring 1 ft in equilibrium. It is attached to a dashpot with constant \(c=12\) lb-sec/ft. The weight is initially displaced 8 inches above equilibrium and released from rest. Find its displacement for \(t>0\).

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\(y=-\dst{2\over3}(e^{-8t}-2e^{-4t})\)
A mass of one kg stretches a spring 49 cm in equilibrium. A dashpot attached to the spring supplies a damping force of 4 N for each m/sec of speed. The mass is initially displaced 10 cm above equilibrium and given a downward velocity of 1 m/sec. Find its displacement for \(t>0\).

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\(y=\dst{e^{-2t}\left({1\over10}\cos 4t-{1\over5}\sin4t\right)}\) m
A mass of 100 grams stretches a spring 98 cm in equilibrium. A dashpot attached to the spring supplies a damping force of 600 dynes for each cm/sec of speed. The mass is initially displaced 10 cm above equilibrium and given a downward velocity of 1 m/sec. Find its displacement for \(t>0\).

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\(y=e^{-3t}(10\cos t-70\sin t)\) cm
A 192 lb weight is suspended from a spring with constant \(k=6\) lb/ft and subjected to an external force \(F(t)=8\cos3t\) lb. Find the steady state component of the displacement for \(t>0\) if the medium resists the motion with a force equal to 8 times the speed in ft/sec.

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\(\dst{y_p=-{2\over15}\cos3t+{1\over15}\sin3t}\) ft
A 2 gm mass is attached to a spring with constant 20 dyne/cm. Find the steady state component of the displacement if the mass is subjected to an external force \(F(t)=3\cos4t-5\sin4t\) dynes and a dashpot supplies 4 dynes of damping for each cm/sec of velocity.

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\(y_p=\dst{{11\over100}\cos4t+{27\over100}\sin4t}\) cm
C/G A 96 lb weight is attached to a spring with constant 12 lb/ft. Find and graph the steady state component of the displacement if the mass is subjected to an external force \(F(t)=18\cos t-9\sin t\) lb and a dashpot supplies 24 lb of damping for each ft/sec of velocity.

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\(\dst{y_p={42\over73}\cos t+{39\over73}\sin t}\) ft
A mass of one kg stretches a spring 49 cm in equilibrium. It is attached to a dashpot that supplies a damping force of 4 N for each m/sec of speed. Find the steady state component of its displacement if it’s subjected to an external force \(F(t)=8\sin2t-6\cos2t\) N.

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\(y=\dst{-{1\over2}\cos2t+{1\over4}\sin2t}\) m
A mass \(m\) is suspended from a spring with constant \(k\) and subjected to an external force \(F(t)=\alpha\cos\omega_0t+\beta\sin\omega_0t\), where \(\omega_0\) is the natural frequency of the spring–mass system without damping. Find the steady state component of the displacement if a dashpot with constant \(c\) supplies damping.

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\(\dst{y_p={1\over c\omega_0}(-\beta\cos\omega_0t+ \alpha\sin\omega_0t)}\)
Show that if \(c_1\) and \(c_2\) are not both zero then

\[ y=e^{r_1t}(c_1+c_2t) \]

can’t equal zero for more than one value of \(t\).
Show that if \(c_1\) and \(c_2\) are not both zero then

\[ y=c_1e^{r_1t}+c_2e^{r_2t} \]

can’t equal zero for more than one value of \(t\).
Find the solution of the initial value problem

\[ my''+cy'+ky=0,\quad y(0)=y_0,\;y'(0)=v_0, \]

given that the motion is underdamped, so the general solution of the equation is

\[ y=e^{-ct/2m}(c_1\cos\omega_1t+c_2\sin\omega_1t). \]

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\(y=\dst{e^{-ct/2m} \left(y_0\cos\omega_1t+{1\over\omega_1}(v_0+{cy_0\over 2m})\sin\omega_1t\right)}\)
Find the solution of the initial value problem

\[ my''+cy'+ky=0,\quad y(0)=y_0,\;y'(0)=v_0, \]

given that the motion is overdamped, so the general solution of the equation is

\[ y=c_1e^{r_1t}+c_2e^{r_2t}\;(r_1,r_2<0). \]

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\(y=\dst{{r_2y_0-v_0\over r_2-r_1} e^{r_1t}+{v_0-r_1y_0\over r_2-r_1}e^{r_2t}}\)
Find the solution of the initial value problem

\[ my''+cy'+ky=0,\quad y(0)=y_0,\;y'(0)=v_0, \]

given that the motion is critically damped, so that the general solution of the equation is of the form

\[ y=e^{r_1t}(c_1+c_2t)\,(r_1<0). \]

Show answer

\(y=e^{r_1t}\dst{\left(y_0+(v_0-r_1y_0)t\right)}\)