HW 5

The block of mass \(m\) will leave the ﬂoor of the box when the vertical acceleration is large enough to match the gravity acceleration \(g\). The equation of motion of the overall system is given by

\begin{equation} y^{\prime \prime }+\omega _{0}^{2}y=0 \tag {1}\end{equation}

\begin{equation} y=A\cos \omega _{0}t+B\sin \omega _{0}t \tag {2}\end{equation}

Initial conditions are used to ﬁnd \(A,B\). Since at \(t=0\), \(y\left ( 0\right ) =d\), then from (2) we ﬁnd

\begin{equation} y^{\prime }=-A\omega _{0}\sin \omega _{0}t+B\omega _{0}\cos \omega _{0}t \tag {3}\end{equation}

At \(t=0\), \(y^{\prime }\left ( 0\right ) =0\), this gives \(B=0\). Therefore the full solution (2) becomes

\begin{align*} y^{\prime } & =-\omega _{0}d\sin \omega _{0}t\\ y^{\prime \prime } & =-\omega _{0}^{2}d\cos \omega _{0}t \end{align*}

The period is \(T_{p}=\frac {2\pi }{\omega _{0}}\). After one \(T_{p}\) from release the box will be the top. Therefore, the acceleration at that moment is

\begin{align*} y^{\prime \prime }\left ( T_{p}\right ) & =-\omega _{0}^{2}d\cos \omega _{0}T_{p}\\ & =-\omega _{0}^{2}d\cos 2\pi \\ & =\omega _{0}^{2}d \end{align*}

The condition for \(m\) to just leave the ﬂoor of the box is when the above acceleration is the same as \(g\).

\begin{align*} \omega _{0}^{2}d & =g\\ d & =\frac {g}{\omega _{0}^{2}}\end{align*}

4.5.2 Problem 2

4.5.2.1 Part (1)

The function \(f\left ( t\right ) \) is an odd function, therefore we only need to evaluate \(b_{n}\) terms. To more clearly see the period, the deﬁnition of \(f\left ( t\right ) \) is written as

\begin{align*} b_{n} & =\frac {1}{\frac {T_{p}}{2}}\int _{0}^{T_{p}}f\left ( t\right ) \sin \left ( n\omega t\right ) dt\\ & =\frac {2}{\frac {2\pi }{\omega }}\left ( \int _{0}^{\frac {\pi }{\omega }}\left ( +1\right ) \sin \left ( n\omega t\right ) dt+\int _{\frac {\pi }{\omega }}^{\frac {2\pi }{\omega }}\left ( -1\right ) \sin \left ( n\omega t\right ) dt\right ) \\ & =\frac {\omega }{\pi }\left ( \int _{0}^{\frac {\pi }{\omega }}\sin \left ( n\omega t\right ) dt-\int _{\frac {\pi }{\omega }}^{\frac {2\pi }{\omega }}\sin \left ( n\omega t\right ) dt\right ) \\ & =\frac {\omega }{\pi }\left ( \left [ -\frac {\cos \left ( n\omega t\right ) }{n\omega }\right ] _{0}^{\frac {\pi }{\omega }}-\left [ -\frac {\cos \left ( n\omega t\right ) }{n\omega }\right ] _{\frac {\pi }{\omega }}^{\frac {2\pi }{\omega }}\right ) \\ & =\frac {\omega }{\pi }\left ( -\frac {1}{n\omega }\left [ \cos \left ( n\omega t\right ) \right ] _{0}^{\frac {\pi }{\omega }}+\frac {1}{n\omega }\left [ \cos \left ( n\omega t\right ) \right ] _{\frac {\pi }{\omega }}^{\frac {2\pi }{\omega }}\right ) \\ & =\frac {1}{n\pi }\left ( -\left [ \cos \left ( n\omega \frac {\pi }{\omega }\right ) -\cos \left ( 0\right ) \right ] +\left [ \cos \left ( n\omega \frac {2\pi }{\omega }\right ) -\cos \left ( n\omega \frac {\pi }{\omega }\right ) \right ] \right ) \\ & =\frac {1}{n\pi }\left ( -\left [ \cos \left ( n\pi \right ) -1\right ] +\left [ \cos \left ( 2n\pi \right ) -\cos \left ( n\pi \right ) \right ] \right ) \\ & =\frac {1}{n\pi }\left ( -\cos \left ( n\pi \right ) +1+\cos \left ( 2n\pi \right ) -\cos \left ( n\pi \right ) \right ) \\ & =\frac {1}{n\pi }\left ( -2\cos \left ( n\pi \right ) +\overset {1}{\overbrace {\cos \left ( 2n\pi \right ) }}+1\right ) \\ & =\frac {2}{n\pi }\left ( 1-\cos \left ( n\pi \right ) \right ) \end{align*}

And since \(n\) is an integer, then \(\cos \left ( n\pi \right ) =\left ( -1\right ) ^{n}\) and the above reduces to

\begin{align*} f\left ( t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\sin \left ( \omega nt\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }\frac {4}{n\pi }\sin \left ( \omega nt\right ) \end{align*}

4.5.2.2 Part (2)

When the system is driven by the above periodic square wave of amplitude \(F_{0}\), the steady state response is the sum to the response of each harmonic in the Fourier series expansion of the forcing function. Since the steady state response of a second order system to \(F_{n}\sin \left ( n\omega t\right ) \) is given by

Then the steady state response to \(f\left ( t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }F_{0}\frac {4}{n\pi }\sin \left ( \omega nt\right ) \) is given by

\begin{align} y_{ss}\left ( t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }\frac {4}{n\pi }\frac {F_{0}/m}{\sqrt {\left ( \omega _{0}^{2}-\left ( n\omega \right ) ^{2}\right ) ^{2}+4\lambda ^{2}\left ( n\omega \right ) ^{2}}}\sin \left ( n\omega t+\delta _{n}\right ) \tag {1}\\ & =\frac {4F_{0}}{\pi m}\sum _{n=1,3,5,\cdots }^{\infty }\frac {1}{n}\frac {\sin \left ( n\omega t+\delta _{n}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\left ( n\omega \right ) ^{2}\right ) ^{2}+4\lambda ^{2}\left ( n\omega \right ) ^{2}}}\nonumber \end{align}

\begin{equation} y_{ss}\left ( t\right ) =\frac {4F_{0}}{\pi m}\left \{ \frac {\sin \left ( \omega t+\delta _{1}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\omega ^{2}\right ) ^{2}+4\lambda ^{2}\omega ^{2}}}+\frac {1}{3}\frac {\sin \left ( 3\omega t+\delta _{3}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\left ( 3\omega \right ) ^{2}\right ) ^{2}+4\lambda ^{2}\left ( 3\omega \right ) ^{2}}}+\frac {1}{5}\frac {\sin \left ( 5\omega t+\delta _{5}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\left ( 5\omega \right ) ^{2}\right ) ^{2}+4\lambda ^{2}\left ( 5\omega \right ) ^{2}}}+\cdots \right \} \tag {2}\end{equation}

We are told that \(3\omega =\omega _{0}\) or \(\omega =\frac {1}{3}\omega _{0}\) and in addition, using using \(Q=\frac {\omega _{0}}{2\lambda }\) we ﬁnd

\begin{align*} 100 & =\frac {\omega _{0}}{2\lambda }\\ \lambda & =\frac {\omega _{0}}{200}\end{align*}

Using this \(\lambda \) and given value of \(\omega \) then the phase \(\delta _{n}\) becomes

\begin{align*} \delta _{n} & =\tan ^{-1}\frac {-2\lambda \left ( n\omega \right ) }{\omega _{0}^{2}-\left ( n\omega \right ) ^{2}}\\ & =\tan ^{-1}\frac {-2\left ( \frac {\omega _{0}}{200}\right ) \left ( n\frac {\omega _{0}}{3}\right ) }{\omega _{0}^{2}-\left ( n\frac {\omega _{0}}{3}\right ) ^{2}}\\ & =\tan ^{-1}\frac {3n}{100n^{2}-900}\end{align*}

\begin{align*} y_{ss}\left ( t\right ) & =\frac {4F_{0}}{\pi m}\left \{ \frac {\sin \left ( \frac {\omega _{0}}{3}t+\tan ^{-1}\frac {-3}{800}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\left ( \frac {\omega _{0}}{3}\right ) ^{2}\right ) ^{2}+4\left ( \frac {\omega _{0}}{200}\right ) ^{2}\left ( \frac {\omega _{0}}{3}\right ) ^{2}}}+\frac {\frac {1}{3}\sin \left ( \omega _{0}t+\frac {\pi }{2}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\left ( 3\frac {\omega _{0}}{3}\right ) ^{2}\right ) ^{2}+4\left ( \frac {\omega _{0}}{200}\right ) ^{2}\left ( 3\frac {\omega _{0}}{3}\right ) ^{2}}}+\frac {\frac {1}{5}\sin \left ( 5\frac {\omega _{0}}{3}t+\tan ^{-1}\frac {3}{320}\right ) }{\sqrt {\left ( \omega _{0}^{2}-\left ( 5\frac {\omega _{0}}{3}\right ) ^{2}\right ) ^{2}+4\left ( \frac {\omega _{0}}{200}\right ) ^{2}\left ( 5\frac {\omega _{0}}{3}\right ) ^{2}}}+\cdots \right \} \\ & =\frac {4}{\pi }\frac {F_{0}}{m}\left \{ \frac {\sin \left ( \frac {\omega _{0}}{3}t-\tan ^{-1}\frac {3}{800}\right ) }{\sqrt {\frac {640\,009}{810\,000}\omega _{0}^{4}}}+\frac {1}{3}\frac {\sin \left ( \omega _{0}t+\frac {\pi }{2}\right ) }{\sqrt {\frac {1}{10\,000}\omega _{0}^{4}}}+\frac {1}{5}\frac {\sin \left ( 5\frac {\omega _{0}}{3}t+\tan ^{-1}\frac {3}{320}\right ) }{\sqrt {\frac {102\,409}{32\,400}\omega _{0}^{4}}}+\cdots \right \} \\ & =\frac {4}{\pi }\frac {F_{0}}{m}\left \{ 1.125\,\frac {\sin \left ( 0.333\omega _{0}t-\tan ^{-1}\frac {3}{800}\right ) }{\omega _{0}^{2}}+33.333\frac {\sin \left ( \omega _{0}t+\frac {\pi }{2}\right ) }{\omega _{0}^{2}}+0.11249\frac {\sin \left ( 1.6667\omega _{0}t+\tan ^{-1}\frac {3}{320}\right ) }{\omega _{0}^{2}}+\cdots \right \} \end{align*}

We see that the third harmonic \(\left ( n=3\right ) \) has the largest amplitude, since this is where \(3\omega =\omega _{0}\). In normalized size, dividing all amplitudes by the smallest amplitude gives

4.5.3 Problem 3

The total force on the planet \(m\) is due to the mass inside the region centered at the center of the sun. The mass outside can be ignored since its eﬀect cancels out. Let the radius of the sun be \(R_{sun}\), then the total mass that pulls the planet toward the center of the solar system is given by

\begin{align*} \bar {F} & =-\frac {GM_{total}m}{r^{2}}\hat {r}\\ & =-\frac {G\left ( M_{sun}+\frac {4}{3}\pi \left ( r^{3}-R_{sun}^{3}\right ) \rho \right ) m}{r^{2}}\hat {r}\end{align*}

Where \(\hat {r}\) is a unit vector pointing from the sun towards the planet \(m\) and \(G\) is the gravitational constant and \(\rho \) is the cloud density.

4.5.4 Problem 4

4.5.4.1 Part (1)

The force on the satellite is \(mr_{e}\omega ^{2}\) where \(r_{e}\) is taken as the earth radius since this is low-lying orbit. Therefore

But \(v=r_{e}\omega \) where \(v\) is the satellite speed we want to ﬁnd. Hence \(\omega ^{2}=\frac {v^{2}}{r_{e}^{2}}\) and the above becomes

\begin{align*} \frac {GM_{e}}{r_{e}^{2}} & =r_{e}\frac {v^{2}}{r_{e}^{2}}\\ v & =\sqrt {\frac {GM_{e}}{r_{e}}}\\ & =\sqrt {\frac {\left ( 6.67408\times 10^{-11}\right ) \left ( 5.972\times 10^{24}\right ) }{6.371\times 10^{6}}}\\ & =7909\,.6\text { meter/sec}\\ & =7.9\text { km/sec}\end{align*}

4.5.4.2 Part (2)

\begin{align*} \frac {GM_{e}m}{r^{2}} & =mr\omega ^{2}\\ r & =\left ( \frac {GM_{e}}{\omega ^{2}}\right ) ^{\frac {1}{3}}\end{align*}

where \(\omega =\frac {2\pi }{T_{p}}\) where \(T_{p}\) is the period of the satellite. But for synchronous satellite, this period is \(24\) hrs. Hence the above becomes

\begin{align*} r & =\left ( \frac {GM_{e}}{\left ( \frac {2\pi }{T_{p}}\right ) ^{2}}\right ) ^{\frac {1}{3}}\\ & =\left ( \frac {\left ( 6.67408\times 10^{-11}\right ) \left ( 5.972\times 10^{24}\right ) }{\left ( \frac {2\pi }{24\left ( 60\right ) \left ( 60\right ) }\right ) ^{2}}\right ) ^{\frac {1}{3}}\\ & =4.224\times 10^{7}\text { meter}\end{align*}

4.5.4.3 Part (3)

\begin{align*} \frac {GM_{e}m}{r^{2}} & =mr\omega ^{2}\\ \frac {GM_{e}}{r^{3}} & =\omega ^{2}\\ \frac {GM_{e}}{r^{3}} & =\left ( \frac {2\pi }{T_{p}}\right ) ^{2}\end{align*}

\begin{align*} \frac {2\pi }{T_{p}} & =\sqrt {\frac {GM_{e}}{r^{3}}}\\ T_{p} & =\frac {2\pi }{\sqrt {\frac {GM_{e}}{r^{3}}}}=\frac {2\pi }{\sqrt {\frac {\left ( 6.67408\times 10^{-11}\right ) \left ( 5.972\times 10^{24}\right ) }{\left ( \left ( 60.3\right ) \left ( 6.371\times 10^{6}\right ) \right ) ^{3}}}}\\ & =2.369\,8\times 10^{6}\text { sec}\end{align*}

\begin{align*} T_{p} & =\frac {2.369\,8\times 10^{6}}{\left ( 24\right ) \left ( 60\right ) \left ( 60\right ) }\\ & =27.428\text { days}\end{align*}

4.5.5 Problem 5

4.5.5.1 Part(1)

From the above diagram, where we have two particles of same mass \(m\) in two circular orbits. The area of each sector is given by

\begin{equation} \frac {dA_{1}}{dt}=\frac {\dot {\theta }_{1}}{2}r_{1}^{2} \tag {1}\end{equation}

\begin{equation} \frac {dA_{2}}{dt}=\frac {\dot {\theta }_{2}}{2}r_{2}^{2} \tag {2}\end{equation}

Since we have a central force, then this force attracts each mass with a force given by \(f=mr\dot {\theta }^{2}\). Therefore \(f_{r_{1}}=mr_{1}\dot {\theta }_{1}^{2}\), Similarly \(f_{r_{2}}=mr_{2}\dot {\theta }_{2}^{2}\). Substituting for \(\dot {\theta }\) from these expressions back into (1) and (2) gives

\begin{equation} \frac {dA_{1}}{dt}=\sqrt {\frac {f_{1}}{mr_{1}}}\frac {r_{1}^{2}}{2} \tag {1B}\end{equation}

\begin{equation} \frac {dA_{2}}{dt}=\sqrt {\frac {f_{1}}{mr_{1}}}\frac {r_{1}^{2}}{2} \tag {2B}\end{equation}

\begin{align*} \frac {dA_{1}}{dt} & =\frac {dA_{2}}{dt}\\ \sqrt {\frac {f_{1}}{mr_{1}}}\frac {r_{1}^{2}}{2} & =\sqrt {\frac {f_{2}}{mr_{2}}}\frac {r_{1}^{2}}{2}\\ \frac {f_{1}}{mr_{1}}\frac {r_{1}^{4}}{4} & =\frac {f_{2}}{mr_{2}}\frac {r_{2}^{2}}{4}\\ f_{1}r_{1}^{3} & =f_{2}r_{2}^{3}\end{align*}

This says that, since we using the same mass, that the force \(f\left ( r\right ) \) on a mass is inversely proportional to the cube of the mass distance from the center. To see this more clearly, let \(r_{1}=1\) then

So if we move the mass from \(r_{1}=1\) to say 3 times as far to \(r_{2}=3\), then the force on the same mass becomes \(\frac {1}{27}\) smaller than it was.

4.5.5.2 Part(2)

\begin{equation} \frac {d^{2}}{d\theta ^{2}}\left ( \frac {1}{r}\right ) +\frac {1}{r}=-\frac {\mu r^{2}}{l^{2}}F\left ( r\right ) \tag {1}\end{equation}

Where \(\mu \) is the reduces mass, \(l\) is the angular momentum and \(F\left ( r\right ) \) is the force we are solving for. Since \(r=r_{0}\cos \theta \) then

\begin{align} \frac {d^{2}}{d\theta ^{2}}\left ( \frac {1}{r}\right ) & =\frac {d}{d\theta }\left ( \frac {d}{d\theta }\frac {1}{r}\right ) =\frac {d}{d\theta }\left ( \frac {d}{d\theta }\frac {1}{r_{0}\cos \theta }\right ) \nonumber \\ & =\frac {d}{d\theta }\left ( \frac {\left ( -1\right ) \left ( -\sin \theta \right ) }{r_{0}\cos ^{2}\theta }\right ) \nonumber \\ & =\frac {d}{d\theta }\left ( \frac {\sin \theta }{r_{0}\cos ^{2}\theta }\right ) \nonumber \\ & =\left ( \frac {\cos \theta }{r_{0}\cos ^{2}\theta }+\frac {2\sin ^{2}\theta }{r_{0}\cos ^{3}\theta }\right ) \nonumber \\ & =\left ( \frac {1}{r_{0}\cos \theta }+\frac {2\sin ^{2}\theta }{r_{0}\cos ^{3}\theta }\right ) \tag {2}\end{align}

But from \(r=r_{0}\cos \theta \) we see that \(\cos \theta =\frac {r}{r_{0}}\) and \(\sin ^{2}\theta =1-\cos ^{2}\theta =1-\left ( \frac {r}{r_{0}}\right ) ^{2}\), hence (2) becomes

\begin{align*} \frac {d^{2}}{d\theta ^{2}}\left ( \frac {1}{r}\right ) & =\left ( \frac {1}{r_{0}\left ( \frac {r}{r_{0}}\right ) }+\frac {2\left ( 1-\left ( \frac {r}{r_{0}}\right ) ^{2}\right ) }{r_{0}\left ( \frac {r}{r_{0}}\right ) ^{3}}\right ) \\ & =\left ( \frac {1}{r}+\frac {2\left ( 1-\frac {r^{2}}{r_{0}^{2}}\right ) }{\frac {r^{3}}{r_{0}^{2}}}\right ) \\ & =\left ( \frac {1}{r}+\frac {2-\frac {2r^{2}}{r_{0}^{2}}}{\frac {r^{3}}{r_{0}^{2}}}\right ) \\ & =\left ( \frac {1}{r}+\frac {2r_{0}^{2}-2r^{2}}{r^{3}}\right ) \\ & =\frac {r^{2}+2r_{0}^{2}-2r^{2}}{r^{3}}\\ & =\frac {2r_{0}^{2}-r^{2}}{r^{3}}\end{align*}