We assume the cube is displaced downwards from its static equilibrium position and it is currently at distance \(x\) below the static position.
The buoyant force \(F_{b}\) will push the cube upwards. This force will equal the weight of water displaced which is \(x a^{2} \rho _{w} g\) where \(\rho _{w}\) is density of water and \(g\) is the gravitational constant. The free body diagram is
Applying \(F=mx^{\prime \prime }\) we obtain equation of motion \begin {align} Mx^{\prime \prime } & = -F_{b}\\ Mx^{\prime \prime }+F_{b} & = 0 \end {align}
\(M=a^{3}\rho \) where \(\rho \) is density of pine. The above can be simplified to\begin {align} a^{3}\rho x^{\prime \prime }+xa^{2}\rho _{w}g & =0\\ x^{\prime \prime }+\frac {\rho _{w}g}{a\rho }x & =0\\ x^{\prime \prime }+\omega _{n}^{2}x & =0 \end {align}
Hence from the above equation \[ \omega _{n}=\sqrt {\frac {\rho _{w}g}{a\rho }}\]
Given \(\rho =400\) \(kg/m^{3}\) and \(\rho _{w}=1000\) \(kg/m^{3}\) and \(a=0.1{m}\) then\[ \omega _{n}=\sqrt {\frac {\rho _{w}g}{a\rho }}=\sqrt {\frac {1000\times 9.81}{0.1\times 400}}=15.66\ \frac {rad}{\sec ^{2}}\] Hence frequency in Hz is\[ f=\frac {\omega _{n}}{2\pi }=\frac {15.66}{2\pi }=2.492\ {hz}\]
applying \(F=mq^{\prime \prime }\), we obtain equation of motion \begin {align} Mq^{\prime \prime } & =-kq\\ Mq^{\prime \prime }+kq & =0\\ q^{\prime \prime }+\frac {k}{M}q & =0\\ q^{\prime \prime }+\omega _{n}^{2}q & =0 \end {align}
Let solution be \(q\relax (t) =\operatorname {Re}\left (\hat {A}e^{i\omega _{n}t}\right ) \) where \(\hat {A}\) is the complex amplitude, which is a complex number that can be written as \(\hat {A}=a+ib\). We use initial conditions to determine \(\hat {A}\). At \(t=0\), let \(q\relax (0) =q_{0}\)\begin {align} q_{0} & =\operatorname {Re}\left (\hat {A}e^{i\omega _{n}t}\right ) \\ & =\operatorname {Re}\left (\hat {A}\right ) \\ & =a \end {align}
Hence \(a=q_{0}\) And since \(q^{\prime }\relax (t) =\operatorname {Re}\left (i\omega _{n}\hat {A}e^{i\omega _{n}t}\right ) \), then \(t=0\) we have\begin {align} q_{0}^{\prime } & =\operatorname {Re}\left (i\omega _{n}\hat {A}\right ) \\ & =\operatorname {Re}\left (i\omega _{n}\left (a+ib\right ) \right ) \\ & =\operatorname {Re}\left (i\omega _{n}a-\omega _{n}b\right ) \\ & =-\omega _{n}b \end {align}
Hence \(b=-\frac {q_{0}^{\prime }}{\omega _{n}}\) therefore the general solution is\begin {align} q\relax (t) & =\operatorname {Re}\left (\hat {A}e^{i\omega _{n}t}\right ) \\ & =\operatorname {Re}\left (\left (a+ib\right ) e^{i\omega _{n}t}\right ) \\ & =\operatorname {Re}\left (\overset {\hat {A}}{\overbrace {\left (q_{0}-i\frac {q_{0}^{\prime }}{\omega _{n}}\right ) }}e^{i\omega _{n}t}\right ) \end {align}
Hence \(\left \vert \hat {A}\right \vert =\sqrt {q_{0}^{2}+\left (\frac {q_{0}^{\prime }}{\omega _{n}}\right ) ^{2}}\) and \(\arg \left (\hat {A}\right ) =\theta =\tan ^{-1}\left (\frac {\frac {q_{0}^{\prime }}{\omega _{n}}}{q_{0}}\right ) \). We have 2 complex quantities above being multiplied. The first is \(\hat {A}\) and the second is \(e^{i\omega _{n}t}\), therefore the result is obtained by adding the angles and by multiplied the magnitudes. The magnitude of \(e^{i\omega _{n}t}\) is one. Hence on the complex plan, the above expression for \(q\relax (t) \) is represented as vector of length \(\left \vert \hat {A}\right \vert \) and phase \(\phi =\theta +\omega _{n}t\)
From the above diagram we see that the maximum value of \[ q_{\max }\relax (t) =\left \vert \hat {A}\right \vert \] which occurs when \[ \phi =\theta +\omega _{n}t=0 \] solving for \(t\) gives\[ t=\frac {-\theta }{\omega _{n}}\] Notice that \(\theta \) is negative, hence we will get positive value for \(t\). Substituting the numerical values given we find thatAnd the earliest time this occurs is\[ t=\frac {1.\allowbreak 3724}{2\pi \left (80\right ) }=2.\allowbreak 7303\times 10^{-3}=2.73\ \text {ms}\] We confirm this by noticing that the initial position vector was at about \(\frac {1}{4}cycle\) away from the positive x-axis (we found the phase of \(\hat {A}\) above to be about \(-80\) degrees), and the rotational speed is given as \(80\) cycles per second. Hence it takes \(12.5\) ms to make one cycle and \(\frac {1}{4}\) of this is about \(3ms\).
Since we found\(\ q\relax (t) =\) \(\operatorname {Re}\left (\left ( q_{0}-i\frac {q_{0}^{\prime }}{\omega _{n}}\right ) e^{i\omega _{n}t}\right ) \), then
\begin {align} q^{\prime }\relax (t) &=\operatorname {Re}\left (i\omega _{n}\left (q_{0}-i\frac {q_{0}^{\prime }}{\omega _{n}}\right ) e^{i\omega _{n}t}\right ) \\ &=\operatorname {Re}\left (\left (\omega _{n}q_{0}-e^{i\frac {\pi }{2}} q_{0}^{\prime }\right ) e^{i\frac {\pi }{2}}e^{i\omega _{n}t}\right ) \\ &=\operatorname {Re}\left (\left (\omega _{n}q_{0}e^{i\frac {\pi }{2}}-e^{i\pi }q_{0}^{\prime }\right ) e^{i\omega _{n}t}\right ) \\ &=\operatorname {Re}\left (\overbrace {\left (\omega _{n}q_{0}e^{i\frac {\pi }{2}}+q_{0}^{\prime }\right ) }^{\hat {B}} e^{i\omega _{n}t}\right ) \end {align}
Where now \(\hat {B}\) is the complex amplitude of \(q^{\prime }\relax (t) \). Hence \(\left \vert \hat {B}\right \vert =\sqrt {\left (\omega _{n}q_{0}\right ) ^{2}+\left (q_{0}^{\prime }\right ) ^{2}}\) and its phase is \(\arg \left ( \hat {B}\right ) =\tan ^{-1}\frac {\omega _{n}q_{0}}{q_{0}^{\prime }}\). The complex plane representation of \(q^{\prime }\relax (t) \) is
From the above diagram we see that maximum magnitude of \(q^{\prime }\left ( t\right ) \) is \(\left \vert \hat {B}\right \vert \) given by\begin {align} \left \vert \hat {B}\right \vert &=\sqrt {\left (\omega _{n}q_{0}\right )^{2}+\left (q_{0}^{\prime }\right ) ^{2}}\\ &=\sqrt {\left (2\pi \left (80\right ) \left (20\times 10^{-3}\right )\right ) ^{2}+\left (-50\right ) ^{2}}\\ &=51.001\text {m/s} \end {align}
The earliest time it occurs is found by solving for \(t\) in\begin {align} \omega _{n}t+\theta &=2\pi \\ t &=\frac {2\pi -\theta }{\omega _{n}}\\ &=\frac {2\pi -\tan ^{-1}\frac {\omega _{n}q_{0}}{q_{0}^{\prime }}}{2\pi \left (80\right ) }=\frac {2\pi -\tan ^{-1}\frac {2\pi \left (80\right ) \left (20\times 10^{-3}\right ) }{-50}}{2\pi \left (80\right ) }\\ =\frac {2\pi -\tan ^{-1}\left (-0.20106\right ) }{2\pi \left (80\right )} &=1.\allowbreak 2895\times 10^{-2}\\ &=0.129\ \text {ms} \end {align}
Adding \(2\) kg caused deflection of \(50\) mm, hence from \(F=k\Delta \) we can find \(k\) as follows\begin {equation} k=\frac {F}{\Delta }=\frac {2g}{0.05}=\frac {2\left (9.81\right ) }{0.05}=\allowbreak 392\ \text {N/m}\label {eq:3.1} \end {equation} where \(g\) is the gravitational constant. We also told that \(f_{2}=f_{1}-5\) where \(f_{2}\) is the natural frequency after adding the second mass and where \(f_{1}=\frac {1}{2\pi }\omega _{1}\) and \(f_{2}=\frac {1}{2\pi }\omega _{2}\), hence\begin {align} f_{2} & =f_{1}-5\\ \frac {1}{2\pi }\omega _{2} & =\frac {1}{2\pi }\omega _{1}-5\\ \omega _{2} & =\omega _{1}-10\pi \end {align}
But \(\omega _{1}=\sqrt {\frac {k}{m}}\) and \(\omega _{2}=\sqrt {\frac {k}{m+2}}\), hence \[ \sqrt {\frac {k}{m+2}}=\sqrt {\frac {k}{m}}-10\pi \] From Eq ?? the above becomes\[ \sqrt {\frac {392}{m+2}}=\sqrt {\frac {392}{m}}-10\pi \] Solving numerically gives \(m=0.1955 \text {kg}\)
To show that \(x\relax (t) =Be^{\lambda t}\) is solution to the differential equation, we substitute this solution into the LHS of the differential equation and see if we obtain zero.\begin {align} x^{\prime }\relax (t) & =\lambda Be^{\lambda t}=\lambda x\left ( t\right ) \\ x^{\prime \prime }\relax (t) & =\lambda ^{2}Be^{\lambda t}=\lambda ^{2}x\relax (t) \end {align}
Then\begin {align} x^{\prime \prime }+2\zeta \omega _{n}x^{\prime }+\omega _{n}^{2}x & =0\\ \lambda ^{2}x\relax (t) +2\zeta \omega _{n}\lambda x\relax (t) +\omega _{n}^{2}x\relax (t) & =0\\ \left (\lambda ^{2}+2\zeta \omega _{n}\lambda +\omega _{n}^{2}\right ) x\left ( t\right ) & =0 \end {align}
Hence \(x\relax (t) =Be^{\lambda t}\) is a non-trivial solution to the differential equation provided \(\lambda ^{2}+2\zeta \omega _{n}\lambda +\omega _{n}^{2}=0\) since then we obtain \(0=0\).
Now we find \(\lambda \) for the different cases.
The roots of \(\lambda ^{2}+2\zeta \omega _{n}\lambda +\omega _{n}^{2}=0\) are\[ \lambda _{1,2}=-\zeta \omega _{n}\pm \omega _{n}\sqrt {\zeta ^{2}-1}\] For underdamped \(\zeta <1\), hence \(\sqrt {\zeta ^{2}-1}<0\) and we write the above as\begin {align} \lambda _{1,2} & =-\zeta \omega _{n}\pm i\omega _{n}\sqrt {1-\zeta ^{2}}\\ & =-\zeta \omega _{n}\pm i\omega _{d} \end {align}
where \[ \omega _{d}=\omega _{n}\sqrt {1-\zeta ^{2}}\] Let \(\lambda =-\zeta \omega _{n}+i\omega _{d}\) and its complex conjugate \(\lambda ^{\ast }=-\zeta \omega _{n}-i\omega _{d}\), hence the solution is\[ x\relax (t) =B_{1}e^{\lambda t}+B_{2}e^{\lambda ^{\ast }t}\] To obtain a real solution we must have \(B_{1}\) be complex say \(\hat {B}\) and \(B_{2}=\hat {B}^{\ast }\). Hence the above can be written as\begin {align} x\relax (t) & =\hat {B}e^{\lambda t}+\hat {B}^{\ast }e^{\lambda ^{\ast }t}\\ & =2\operatorname {Re}\left (\hat {B}e^{\lambda t}\right ) \\ & =\operatorname {Re}\left (2\hat {B}e^{\left (-\zeta \omega _{n}+i\omega _{d}\right ) t}\right ) \end {align}
Therefore\begin {equation} x\relax (t) =\operatorname {Re}\left (\hat {A}e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}\right ) \label {eq:4.1} \end {equation} Where \(\hat {A}=2\hat {B}=a+ib\). Hence\[ x\relax (t) =\operatorname {Re}\left (\left (a+ib\right ) e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}\right ) \] To find \(a,b\) we need to use initial conditions. Assuming \(x\relax (0) =x_{0}\) and \(x^{\prime }\relax (0) =x_{0}^{\prime }\) then from Eq ?? we obtain\[ x_{0}=\operatorname {Re}\left (a+ib\right ) =a \] Hence\[ a=x_{0}\] and taking derivative of ?? \(\ \)\begin {align} x(t) &=\operatorname {Re}\left (\left (a+ib\right )e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}\right ) \\ x^{\prime }\relax (t) &=\operatorname {Re}\left (-\zeta \omega _{n}\left (a+ib\right ) e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}+i\omega _{d}\left (a+ib\right ) e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}\right )\\ x^{\prime }\relax (0) &=\operatorname {Re}\left (-\zeta \omega _{n}\left (a+ib\right ) +i\omega _{d}\left (a+ib\right ) \right ) \\ &=-\zeta \omega _{n}a-\omega _{d}b \end {align}
Hence \[ b=\frac {x_{0}^{\prime }+\zeta \omega _{n}a}{\omega _{d}}\] But \(a=x_{0}\), hence\[ b=\frac {x_{0}^{\prime }+\zeta \omega _{n}x_{0}}{\omega _{d}}\] Hence ?? becomes\begin {align} x\relax (t) & =\operatorname {Re}\left (\left (a+ib\right ) e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}\right ) \\ & =\operatorname {Re}\left (\left (x_{0}+i\frac {x_{0}^{\prime }+\zeta \omega _{n}x_{0}}{\omega _{d}}\right ) e^{-\zeta \omega _{n}t}e^{i\omega _{d}t}\right ) \end {align}
And this is the general solution. In complex plan it is
Hence the rotating vector will have its length become smaller with time since \(\left \vert \hat {A}\right \vert \) is multiplied by \(e^{-\zeta \omega _{n}t}\). The real part, which is the solution will eventually damp down to zero. Hence it is a damped sinusoid oscillation as follows
From \[ \lambda _{1,2}=-\zeta \omega _{n}\pm \omega _{n}\sqrt {\zeta ^{2}-1}\] For overdamped \(\zeta >1\), hence \(\sqrt {\zeta ^{2}-1}>0\) and we write the above as\[ \lambda _{1,2}=-\zeta \omega _{n}\pm \omega _{n}\sqrt {1-\zeta ^{2}}\] Hence the solution is\[ x\relax (t) =B_{1}e^{\lambda _{1}t}+B_{2}e^{\lambda _{2}t}\] where \(\lambda _{1}=-\zeta \omega _{n}+\omega _{n}\sqrt {1-\zeta ^{2}}\) and \(\lambda _{2}=-\zeta \omega _{n}-\omega _{n}\sqrt {1-\zeta ^{2}}\). We see that both roots are negative always, hence we have 2 exponentially damped solution being added with no oscillation. A sketch of the solution is
From looking at the plot above, here are the values estimated for displacement positive peaks and time they occur
| \(t\) | \(y\relax (t) \) |
| \(0.07\) | \(16\) |
| \(0.17\) | \(12\) |
| \(0.27\) | \(9\) |
| \(0.37\) | \(6\) |
From the above we estimate the natural period \(T \approx 0.1 \sec \) hence \(f=10\ \text {hz}\) hence \(\omega _{n} = 2 \pi f = 60.3\ \text {rad/sec}\) The log decrement is\[ \delta = \ln \frac {y_{i}}{y_{i+N}} \] Select \(i=1\) and \(N=3\) gives \begin {align} \delta & = \ln \frac {16}{6} \\ & = 0.981 \end {align}
To find \(\zeta \) we use the log decrement method \[ \delta =2\pi N \zeta \] Hence\begin {align} \zeta & = \frac {\delta }{2\pi N}=\frac {0.98083}{2\pi \relax (3) }\\ \zeta & = 0.052 \end {align}
Hence\[ \zeta =5.2 \% \]
\[ \ln \left (\frac {y_{1}}{y_{1+N}}\right ) = 2 \pi N \zeta \] Where now we write \(y_{1}=16\) and \(y_{N+1}=0.01\), and hence we need to find \(N\) the only unknown in the equation above\[ \ln \left (\frac {16}{0.01} \right ) = 2 \pi N (0.052) \] Hence\[ N=\frac {\ln \left (\frac {16}{0.01}\right ) }{2\pi \left (0.052\right )} = 22.581 \] We take \(N=23\). What this says is that after 23 periods beyond the first peak, we will satisfy the requirement. But \(T=0.1\) sec, and the first peak was at \(t=0.05\) sec, therefore\begin {align} t &= 0.07+23(0.1) \\ &= 2.37 \text {sec} \end {align}
Since \(\delta _{N} = 2 \pi N \zeta \) and \(\zeta = \frac {c}{c_{r}} = \frac {c}{2 \sqrt {km}} \), then if we double \(k\) and half the mass \(m\), then \(\zeta \) would remain the same since \(c\) is held constant. Therefore the answer in part b would not change.
Since this is an underdamped system, the solution is \[ q\relax (t) =\operatorname {Re}\left (\hat {A}e^{\left (-\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \] Where \(\hat {A}\) is the complex amplitude, say \(\left (a+ib\right ) \), hence\[ q\relax (t) =\operatorname {Re}\left (\left (a+ib\right ) e^{\left ( -\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \] At \(t=0\) we find that \[ a=q\relax (0) =q_{0}\] Hence\[ a = -0.01 \] and the general solution is\[ q\relax (t) =\operatorname {Re}\left (\left (q_{0}+ib\right ) e^{\left (-\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \] Now taking derivatives of the above gives\begin {equation} q^{\prime }\relax (t) =\operatorname {Re}\left (\left (-\omega _{n}\zeta +i\omega _{d}\right ) \left (q_{0}+ib\right ) e^{\left (-\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \nonumber \end {equation} At \(t=0\) then, assuming \(q_{0}^{\prime }\) is the initial velocity\begin {align} q_{0}^{\prime } & =\operatorname {Re}\left (\left (-\omega _{n}\zeta +i\omega _{d}\right ) \left (q_{0}+ib\right ) \right ) \\ & =-\omega _{n}\zeta q_{0}-\omega _{d}b \end {align}
Hence\[ b=-\frac {q_0^\prime +\omega _n\zeta q_0}{\omega _d} \] Therefore the general solution is\[ q\relax (t) =\operatorname {Re}\left (\left (q_{0}-i\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\right ) e^{\left ( -\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \] and \[ q^{\prime }\relax (t) =\operatorname {Re}\left (\left (-\omega _{n}\zeta +i\omega _{d}\right ) \left (q_{0}-i\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\right ) e^{\left (-\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \] Now at \(t_{0}=0.07\) sec the velocity is zero, since this is where the displacement is maximum (first peak). Hence now we have one equation with one unknown \(q_{0}^{\prime }\)that we can solve for from the above\begin {align} 0 &=\operatorname {Re}\left (\left (-\omega _{n}\zeta +i\omega _{d}\right )\left (q_{0}-i\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\right )e^{\left (-\omega _{n}\zeta +i\omega _{d}\right ) t_{0}}\right ) \\ &=e^{-\omega _{n}\zeta t_{0}}\operatorname {Re}\left (\left (-\omega _{n}\zeta q_{0}+i\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{n}\zeta +iq_{0}\omega _{d}+\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{d}\right ) e^{i\omega _{d}t_{0}}\right )\\ &=e^{-\omega _{n}\zeta t_{0}}\operatorname {Re}\left (\left (i\left (\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{n}\zeta +q_{0}\omega _{d}\right ) +q_{0}^{\prime }\right ) e^{i\omega _{d}t_{0}}\right )\\ &=e^{-\omega _{n}\zeta t_{0}}\operatorname {Re}\left (i\left (\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{n}\zeta +q_{0}\omega _{d}\right ) e^{i\omega _{d}t_{0}}+q_{0}^{\prime }e^{i\omega _{d}t_{0}}\right )\\ &=e^{-\omega _{n}\zeta t_{0}}\operatorname {Re}\left (\frac {-1}{i}\left (\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{n}\zeta +q_{0}\omega _{d}\right ) e^{i\left (\omega _{d}t_{0}\right ) }+q_{0}^{\prime }e^{i\omega _{d}t_{0}}\right )\\ &=e^{-\omega _{n}\zeta t_{0}}\left [ \operatorname {Re}\left (\frac {-1}{i}\left (\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{n}\zeta +q_{0}\omega _{d}\right ) e^{i\left (\omega _{d}t_{0}\right )}\right ) +\operatorname {Re}\left (q_{0}^{\prime }e^{i\omega _{d}t_{0}}\right )\right ]\\ &=e^{-\omega _{n}\zeta t_{0}}\left [ -\left (\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\omega _{n}\zeta +q_{0}\omega _{d}\right )\sin \left (\omega _{d}t_{0}\right ) +q_{0}^{\prime }\cos \left (\omega _{d}t_{0}\right ) \right ] \end {align}
But \(q_{0}=-0.01\) m/sec\(,\) \(\zeta =0.052\), \(\omega _{n}=60.3\) rad/sec and \(\omega _{d}=\omega _{n}\sqrt {1-\zeta ^{2}}=60.3\sqrt {1-0.052^{2}}=\allowbreak 60.\allowbreak 218\), therefore \(\omega _{d}t_{0}=\allowbreak 60.\allowbreak 218\times 0.07=4.2153\) and \(q_{0}\omega _{d}=\left (60.218\right ) \left ( -0.01\right ) =-0.60218\ \)hence the above equation becomes\begin {align} 0 &=e^{-\left (60.3\right ) \left (0.052\right ) \left (0.07\right )}\left (-\left (\frac {q_{0}^{\prime }+\left (60.3\right ) \left (0.052\right ) \left (-0.01\right ) }{60.218}60.3\times 0.052-0.602\right )\sin \left (4.215\right ) +q_{0}^{\prime }\cos \left (4.215\right ) \right )\\ &=0.80293\left (-\left (\frac {q_{0}^{\prime }-3.136\times 10^{-2}}{60.218}(3.1356) -0.602\right )\left (-0.879\right ) +q_{0}^{\prime }\left (-0.477\right ) \right ) \end {align}
Solving for \(q_{0}^{\prime }\) gives\[ q_{0}^{\prime }=-1.231\ \text {m/sec}\] Now that we \(q_{0}'\),
Now that we \(q_{0}^{\prime }\), we can find the numerical value for \(b\) and write the general solution again.\begin {align} b & = -\frac {q_{0}^{\prime }+\omega _{n}\zeta q_{0}}{\omega _{d}}\\ & = -\frac {-1.231+60.3\left (0.052\right ) \left (-0.01\right ) }{60.218}\\ & = 2.0963\times 10^{-2} \end {align}
Hence from\begin {align} q\relax (t) & =\operatorname {Re}\left (\left (q_{0}+ib\right ) e^{\left (-\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \\ & =\operatorname {Re}\left (\left (-0.01+i0.0209\right ) e^{\left ( -\omega _{n}\zeta +i\omega _{d}\right ) t}\right ) \end {align}
giving\(\left \vert \hat {A}\right \vert =\sqrt {0.01^{2}+0.0209^{2}}=0.023\)
Assume the system is underdamped.
When the package hits the ground, its speed becomes zero. Therefore the impulse generated on it is the change of linear momentum. Since it speed was \(v\) just before impact, then impulse\(=mv\).
Hence the EQM is\begin {align} mz^{\prime \prime } & = -cz^{\prime }-kz-F_{impulse}+mg \\ mz^{\prime \prime }+cz^{\prime }+kz & = mg-F_{impulse} \end {align}
With the initial conditions now being \(z=0\) and \(z^{\prime }=0\).
The response due to the force \(mg\) can be found from the response to a unit step of amplitude \(mg\) Hence the response due to the force \(mg\) is\[ u(t) = \frac {mg}{k} \left (1-e^{-\zeta \omega _{n}t}\left [\cos \omega _{d}t+\frac {\zeta \omega _{n}}{\omega _{d}}\sin \omega _{d}t\right ] \right ) \] The response due to the impulse is the response of a free system with zero initial position but with initial velocity \(\frac {impulse}{m}\) in the upward (negative) direction. Hence the response due to the impulse only is\begin {align} g\relax (t) & = \frac {mv}{m\omega _{d}}e^{-\zeta \omega _{n}t}\sin \omega _{d}t\\ & = \frac {v}{\omega _{d}}e^{-\zeta \omega _{n}t}\sin \omega _{d}t \end {align}
Hence the downward displacement is given by\begin {align} z(t) = \frac {mg}{k}\left (1-e^{-\zeta \omega _{n}t}\left [ \cos \omega _{d}t+\frac {\zeta \omega _{n}}{\omega _{d}}\sin \omega _{d}t\right ] \right ) -\frac {ve^{-\zeta \omega _{n}t}}{\omega _{d}}\sin \omega _{d}t \end {align}
Now that the impulse have taken place and we have accounted for it in the \(z\relax (t) \) solution, then we can use this expression to find the spring force since \(F_{spring}=kz\relax (t) \) and the damping force on the mass \(F_{damper}=cz^{\prime }\relax (t) \). When resultant net force \(F\) is negative then the mass will rebound from the ground.
\[ F = mg-kz(t) -cz^{\prime }(t) \] But \begin {align} z\relax (t) &=\frac {mg}{k}\left (1-e^{-\zeta \omega _{n}t}\left [ \cos \omega _{d}t+\frac {\zeta \omega _{n}}{\omega _{d}}\sin \omega _{d}t\right ] \right ) -\frac {ve^{-\zeta \omega _{n}t}}{\omega _{d}}\sin \omega _{d}t \end {align}
Hence \begin {align} z^{\prime }\relax (t) &=\frac {mg}{k}\left (\zeta \omega _{n} e^{-\zeta \omega _{n}t}\left [ \cos \omega _{d}t+\frac {\zeta \omega _{n}}{\omega _{d}}\sin \omega _{d}t\right ] -e^{-\zeta \omega _{n}t}\left [ -\omega _{d}\sin \omega _{d}t+\zeta \omega _{n}\cos \omega _{d}t\right ] \right ) -\frac {\zeta \omega _{n}ve^{-\zeta \omega _{n}t}}{\omega _{d}}\sin \omega _{d}t-ve^{-\zeta \omega _{n}t}\cos \omega _{d}t \end {align}
or\begin {align} z^{\prime }\relax (t) &=\frac {e^{-\zeta \omega _{n}t}}{k\omega _{d}}\left ( \left (kv\omega _{n}\zeta +gm\left (\omega _{d}^{2}+\omega _{n}^{2}\zeta ^{2}\right ) \right ) \sin \omega _{d}t-kv\omega _{d}\cos \left (\omega _{d}t\right ) \right ) \end {align}
Hence\begin {align} F &=mg-kz\relax (t) -cz^{\prime }\relax (t) \\ &=mg-k\left [ \frac {mg}{k}\left (1-e^{-\zeta \omega _{n}t}\left [ \cos \omega _{d}t+\frac {\zeta \omega _{n}}{\omega _{d}}\sin \omega _{d}t\right ] \right )-\frac {ve^{-\zeta \omega _{n}t}}{\omega _{d}}\sin \omega _{d}t\right ] -c\frac {e^{-\zeta \omega _{n}t}}{k\omega _{d}}\left (\left (kv\omega _{n}\zeta +gm\left (\omega _{d}^{2}+\omega _{n}^{2}\zeta ^{2}\right ) \right )\sin \omega _{d}t-kv\omega _{d}\cos \left (\omega _{d}t\right ) \right ) \end {align}
To find when this force will turn negative first time, we can take the derivative with respect to time and set it to zero and solve for first \(t=t^{\prime }\) that will make it zero. Since the force was positive first, then it has to become zero before turning negative.
Let \(m=1\) kg, \(\omega _{n}=5\) rad/sec3, \(v=4\) m/s\(.\) Hence \(\omega _{d}=5\sqrt {1-\zeta ^{2}}\). Since \(\omega _{n}^{2}=\frac {k}{m}\), hence \(k=25\) N/m. Also \(c=\zeta c_{cr}=\zeta 2m\omega _{n}=10\zeta \)
Using these values, the force in part(b) is plotted for different values of \(\zeta \). For example, setting \(\zeta =5\%\) gives this plot of \(F\relax (t)\) for \(t=0\) to \(20\) seconds.
The maximum force is seen as little over \(20\) N.Therefore, to find which \(\zeta \) gives the smallest value of maximum force, we can try different values of \(\zeta \) and see how the maximum force changes as a function of \(\zeta \). Using software the following values of maximum for for different \(\zeta \) are generated along with \(t=t_{\max }\) when this maximum occurs and with the time \(t=t^{\prime }\) when the mass rebounds first time from \(z=0\)
| maximum force (N) | \(\zeta \%\) | \(t_{\max }(\sec )\) | \(t^{\prime }(\sec )\) |
| \(22.1\) | \(1\) | \(0.21\) | \(0.84\) |
| \(21.85\) | \(3\) | \(0.206\) | \(0.834\) |
| \(21.45\) | \(7\) | \(0.184\) | \(0.81\) |
| \(21.27\) | \(10\) | \(0.16\) | \(0.779\) |
| \(21.4\) | \(20\) | \(0.11\) | \(0.75\) |
| \(25.8\) | \(40\) | \(0.01\) | \(0.68\) |
| \(30\) | \(50\) | \(0.001\) | \(0.64\) |
Most protection when damping ratio is below \(10\%\)
3typo in book. hz is assumed to mean rad/sec