PHYS 35100 - FALL 2024

Homework Set 2 - Solutions

1. Teenagers

   1. Terminal velocity is reached when the drag force is equal to the force of gravity, thus there is no more acceleration: \begin{equation} m \frac{dv}{dt} = m g - c v^2 = 0 \label{eq:newtonianDrag2ndlaw} \end{equation} here, $c$ is the coefficient determined by the various physical properties: \begin{equation} c = \frac{1}{2} \rho c_d A \end{equation} We can express all this in a way to just use the diameters, $D$ of the balls: $$c = \frac{1}{2} \rho c_d \pi \frac{D^2}{4} = 0.22 D^2$$ using $\rho = 1.204 \rm{kg}/\rm{m}^3$ and $c_d = 0.47$. Thus we have: \begin{equation} v_\rm{ter} = \sqrt{ \frac{mg}{0.22 D^2}} \end{equation}

      Bowling Ball	15 lb	6.80389 kg	8.595"	0.218313 m	79.3 m/s
      Basket Ball	x	0.625 kg	9"	0.2286 m	23.0 m/s

   2. From the notes, we obtained an analytical expression for $x(t)$: \begin{equation} y(t) = \frac{v_\rm{ter}^2}{g} \ln \left(\cosh \left(\frac{gt}{v_\rm{ter}} \right) \right) \end{equation} We can rearrange this for $t(y)$ to get the time it takes to fall a distance $h$. \begin{equation} t = \frac{v_\rm{ter}}{g} \textrm{arccosh} \left( e^\left( \frac{h g}{v_\rm{ter}^2}\right)\right) \end{equation} Putting in our numbers we find:

      Bowling Ball	2.24 s
      Basketball	2.40 s

   3. Here is a set of functions that could be used to generate these results:

      
      import numpy as np
      
      g = 9.8
      rho = 1.204
      cd = 0.47
      
      def calc_vterm_imp(m,D):
          gamma = (1/2)*rho*cd*(1/4)*np.pi
          vterm = np.sqrt((m*0.45359237*g) /(gamma*((D*0.0254)**2) ))
          return vterm
      
      def calc_vterm_met(m,D):
          gamma = (1/2)*rho*cd*(1/4)*np.pi
          vterm = np.sqrt((m*g) /(gamma*((D)**2) ))
          return vterm
      
      print(calc_vterm_met(.625,0.2286))
      
      def calc_time(m,D,h):
        t = (calc_vterm_met(m,D)/g)*np.arccosh(np.exp(h * g/(calc_vterm_met(m,D)**2) ))
        return t
      
      print(calc_time(6.80389,0.218313,24.38))
      
      print(calc_time(0.625,0.2286,24.38))
              

2. Both Kinds

   1. From Newton's 2nd Law: \begin{equation} m \overset{\bullet}{v} = - bv - cv^2 \end{equation} Separate the variables: $$ \frac{m dv}{\left(bv + cv^2 \right)} = -dt $$ Doing the integration will lead to: $$\frac{m}{b}\ln \left( \frac{v}{b+ cv} \right) = -t$$ and evaluating between $v_0, v$ and $0,t$. we can reach $$ v = \frac{bAe^{-bt/m}}{1-cAe^{-bt/m}}$$ $A$ is a constant given by the initial conditions: $$A = \frac{v_0}{b+c v_0}$$

   2. Try plotting both. You can see in the short time, the $v(t)$ tracks with the quadratic drag. After long times, it begins following the linear drag function.

      

3. Need to know

   Consider the total energy at two times: \begin{equation} E = \frac{1}{2} m v_1^2 + \frac{1}{2}k x_1^2 \; \;\textrm{and}\; \; E = \frac{1}{2} m v_2^2 + \frac{1}{2}k x_2^2 \end{equation} Solving for $\omega^2 = \frac{k}{m}$ \begin{equation} \omega = \sqrt{ \frac{v_1^2-v_2^2}{x_2^2-x_1^2}} \end{equation}

   Also, $E = \frac{1}{2}kA^2$ thus: \begin{equation} A = \sqrt{ \frac{x_2^2 v_1^2 -x_1^2 v_2^2}{v_1^2 - v_2^2}} \end{equation}

4. Damped

   1. To convert the frequencies (f) to period (T), we simply take the inverse: $f_0 = 1000 \; \rm{Hz} $ and $T = \frac{1}{f}$ so $$ T_0 = \frac{1}{1000 \; \rm{Hz}} = 0.001 \; \rm{s} = 1 \; \rm{ms}$$ and the damped period is $$ T_1 = \frac{1}{989 \; \rm{Hz}} = 0.00101112 \; \rm{s} = 1.011 \; \rm{ms}$$
   2. We can express the damped frequency $\omega_1$ in terms of $\omega_0$ and $\beta$: $$ \omega_1 = \sqrt{\omega_0 - \beta^2} $$ Solving for $\beta$ we can obtain: $$ \beta = \omega_0 \sqrt{1-\frac{\omega_1^2}{\omega_0^2}} $$

      I'll use python to do the calculations

      
      import numpy as np
      import matplotlib.pyplot as plt
      
      f0 = 1000
      f1 = 989
      omega0 = 2*np.pi*f0
      omega1 = 2*np.pi*f1
      
      beta = omega0*np.sqrt(1-(omega1**2/omega0**2))
      print("The value for beta is then: "+str(beta))
      
      

      Which results in a value for $\beta$ of 929.38

   3. To figure out how many oscillations will happen before the amplitude decreases by half, we can use the fact that: $$e^{-\beta t_\rm{half}} = 0.5$$ and solve for $t_\rm{half}$.

                  
      tHalf = np.log(.5)/(-beta)
      print("t_half is then given by: "+str(tHalf) +" s")
                  
              

      This leads to a value for $t_\rm{half}$ : 0.00074582 s

      Knowing our period is around 1 millisecond, we can tell right away that not even one oscillation will occur before the amplitude is decreased by half. Plotting the damped oscillator equation confirms this.