Example 1 The maximum velocity attained by the mass of a simple harmonic oscillator shown in Figure 1 is 10 cm/s, and the period of oscillation is 2 s. If the mass is released with an initial displacement of 2 cm find:
Figure 1 - Simple harmonic oscillator.
Solution The differential equation that describes the motion of the harmonic oscillator can be written as:
\begin{eqnarray} m\ddot{x} + kx &=& 0\nonumber \end{eqnarray} The general solution to previous differential equation can be written as: \begin{eqnarray} x &=& A\cos(\omega_n t - \phi) \end{eqnarray} a) The Amplitude The linear velocity is \(10 \left[\frac{\mathrm{cm}}{\mathrm{s}}\right] = 0.1 \left[\frac{\mathrm{m}}{s}\right]\), and the period of \(2 [\mathrm{s}]\). The amplitude can be found from relation \(v = \omega_n\cdot A \). However the \(\omega_n\) must be calculated from period. \begin{eqnarray} \tau_n &=& \frac{2\pi}{\omega_n}\Big/\cdot\frac{\omega_n}{\tau_n}\nonumber\\ \omega_n &=& \frac{2\pi}{\tau_n} = \frac{2\pi}{2} = 3.1416 \left[\frac{rad}{s}\right]\nonumber \end{eqnarray} Now the amplitude can be calculated: \begin{eqnarray} v &=& \omega_n\cdot A\Big/\cdot \frac{1}{\omega_n}\nonumber\\ A &=& \frac{v}{\omega_n} = \frac{0.1}{3.1416} = 0.03183 \left[\mathrm{m}\right].\nonumber \end{eqnarray} d) the phase angle The body with mass \(m\) is released at initial displacement of \(0.02\left[\mathrm{m}\right]\): \begin{eqnarray} x(t=0) &=& 0.02 \left[\mathrm{m}\right]\nonumber\\ x(0) &=& A\cos(-\phi_0) = 0.02\nonumber\\ \cos(-\phi_0) &=& \frac{0.02}{A} = 0.6283\nonumber\\ \phi_0 &=& 51.0724°\nonumber. \end{eqnarray} b) the initial velocity \begin{eqnarray} \dot{x}_0 &=& \dot{x}(t = 0) = -\omega_nA\sin(-\phi_0)\nonumber\\ \dot{x}-0 &=& -0.1 \sin(-51.0724°) = 0.07779 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\nonumber \end{eqnarray} c) the maximum acceleration - for maximum acceleration the general solution must be derived twice wit respect to time. \begin{eqnarray} \dot{x} &=& -A\omega_n \sin(\omega_n t+\phi_0)\nonumber\\ \ddot{x} &=& -A\omega_n^2 \cos(\omega_n t + \phi_0)\nonumber \end{eqnarray} To calculate the maximum acceleration the trigonometric part in previous equation is neglected nad absolute value of \(-\omega_n^2 A\) is used. Now using the second equation the acceleration is equal to: \begin{eqnarray} \ddot{x} &=& \omega^2 A = (3.1416)^2\cdot(0.03183) = 0.314151 \left[\frac{\mathrm{m}}{\mathrm{s}^2s}\right]\nonumber \end{eqnarray} Example 2 - an automobile having a mass of \(5000 \left[\mathrm{kg}\right]\) deflects its suspension springs \(0.05 \left[\mathrm{m}\right]\), under static conditions. Determine the natural frequency of the automobile in the vertical direction. Damping is negligible. \begin{eqnarray} \omega_n &=& \sqrt{\frac{g}{\delta_{st}}}\nonumber\\ \omega_n &=& \sqrt{\frac{9.81}{0.05}} = 196.2 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} Example 3 A rigid block of mass M is mounted on four elastic supports, as shown in Figure 2. As mass \(m\) drops from a height \(l\) and adheres to the rigid block without rebounding. If the sping of elastic support is \(k\), find the natural freuqnecy of vibration of the system:
Figure 2 - The body of mass \(m\) dropping on a spring-supported rigid block with mass \(M\).
Solution: In case a) without the mass \(m\) the natural frequency can be written as:
\begin{eqnarray}
\omega_n &=& \sqrt{\frac{k_t}{M}} = \sqrt{\frac{4k}{M}}.\nonumber
\end{eqnarray}
The \(k_t\) is the total stiffness in the system which is equal to stifness of all four springs in the system. \(M\) is the mass of the rigid block.
In case b) with mass \(m\) the natural frequency can be written as: \begin{eqnarray} \omega_n &=& \sqrt{\frac{4k}{M+m}}.\nonumber \end{eqnarray} Velocity of the falling mass: \begin{eqnarray} v &=& \sqrt{2gl}\nonumber \end{eqnarray} At static equilibrium postion the \(x = 0\). At \(x_0 (t = 0) = -\frac{mg}{4k}\). Conservation of momentum: \begin{eqnarray} (M+m)\dot{x}_0 &=& mv = m\sqrt{2gl}\nonumber\\ \dot{x}_0 &=& \dot{x}(t=0) = \frac{m}{M+m} \sqrt{2gl}\nonumber \end{eqnarray} Complete solution: \begin{eqnarray} x(t) &=& A_0\sin(\omega_n t+\phi_0)\nonumber\\ A_0 &=& \sqrt{x_0^2 + \left(\frac{\dot{x}_0}{\omega_n}\right)^2}\nonumber\\ A_0 &=& \sqrt{\frac{m^2 g^2}{16k^2} + \frac{m^2 gl}{2k(M+m)}}\nonumber\\ \phi_0 &=& \tan^{-1}\left(\frac{x_0 \omega_n}{\dot{x}_0}\right) = \tan^{-1}\left(\frac{-\sqrt{g}}{\sqrt{glk(M+m)}}\right)\nonumber \end{eqnarray} Example 4 A hammer with mass \(6 \left[\mathrm{kg}\right]\) strikes anvil of mass \(50 \left[\mathrm{kg}\right]\) with a velocity of \(15\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\) as shown in Figure 3. The anvil is uspported on four springs, each of stiffness \(k = 17.5\left[\frac{\mathrm{kN}}{\mathrm{m}}\right]\). find the resulting motion of the anvil a) if the hammer remains in the contact with the anvil and b) if the hammer does not remain in contact with the anvil after the initial impact.
Figure 3 - Hammer striking an anvil.
Solution: The velocity of the hammer is \(v = 15\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\), the \(x=0\) at static equilibrium position. \begin{eqnarray} x_0 &=& x(t=0) = -\frac{\mathrm{weight}}{k_{eq}} = -\frac{mg}{4k}\nonumber \end{eqnarray} Conservation of momentum: \begin{eqnarray} (M+m)\dot{x}_0 &=& mv\nonumber\\ \dot{x}_0 &=& \frac{mv}{M+m}\nonumber \end{eqnarray} Natural frequency: \begin{eqnarray} \omega_n &=& \frac{4k}{M+m}\nonumber \end{eqnarray} Complete solution can be written as: \begin{eqnarray} x(t) &=& A_0 \sin (\omega_n t + \phi_0)\nonumber \end{eqnarray} where: \begin{eqnarray} A_0 &=& \sqrt{x_0^2 + \left(\frac{\dot{x}_0}{\omega_n}\right)^2}\nonumber\\ &=& \sqrt{\frac{m^2 g^2}{16k^2} + \frac{m^2 v^2}{(M+m)4k}}\nonumber\\ \phi_0 &=& \tan^{-1}\left(-\frac{g\sqrt{M+m}}{v\sqrt{4k}}\right)\nonumber \end{eqnarray} \(v= 15\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\), \(m = 6 \left[\mathrm{kg}\right]\), \(M = 50 \left[\mathrm{kg}\right]\), \(k = 17.5 \left[\frac{\mathrm{kN}}{\mathrm{m}}\right]\).
\begin{eqnarray} A_0 &=& \sqrt{\frac{6^2 \cdot 9.81^2 }{16 \cdot 17.5^2} + \frac{6^2 \cdot 15^2 }{(50 + 6)\cdot 4 \cdot 17.5\cdot 10^3}}\nonumber\\ &=& 0.0454646 \left[\mathrm{m}\right]\nonumber\\ \phi_0 &=& \tan^{-1}\left(-\frac{9.81\cdot \sqrt{50+6}}{15\sqrt{4\cdot 17.5\cdot 10^3}}\right)\nonumber\\ &=& -0.0184958\nonumber \end{eqnarray} b) At static equilibrium postion: \(x_0 = x(t=0) = 0\) Conservation of momentum gives: \begin{eqnarray} M\dot{x}_0 &=& mv\nonumber\\ x_0 &=& x(t=0) = \frac{mv}{M}\nonumber\\ x_0 &=& x(t=0) = 0\nonumber \end{eqnarray} The amplitude: \begin{eqnarray} A_0 &=& \frac{mv}{\sqrt{4kM}} = \frac{6\cdot 15}{\sqrt{4\cdot 17.5\cdot 10^3 \cdot 50}}\nonumber\\ A_0 &=& 0.048107 \left[\mathrm{m}\right]\nonumber\\ \phi_0 &=& \tan^{-1}\left(\frac{x_0\omega_n}{\dot{x}_0}\right) = \tan^{-1}(0) = 0 \end{eqnarray} Example 5 A heavy machine weighing 9810 N is being lowered vertically down by a winch at velocity of 2 m/s. The steel cable supporting the machine has a diameter of 0.01 m. The winch is suddenly stopped when the steel cable's length is 20 m. Find the period of amplitude of the ensuing vibration of the machine. Solution: The mass of the machine can be calculated by dividing the machine weight by gravitational acceleration \(9.81\left[\frac{\mathrm{m}}{\mathrm{s}^2}\right]\) i.e: \begin{eqnarray} m &=& \frac{G}{g} =\frac{9810}{9.81} = 1000 \left[\mathrm{kg}\right]\nonumber \end{eqnarray}
To calculate the natural frequency first the stiffness must be calculated. The diameter of the cable is \(0.01 \left[\mathrm{m}\right]\), and the cable is made from steel so the modulus of elasticity is equal to \(2.07\cdot 10^{11} \left[\mathrm{Pa}\right]\) \begin{eqnarray} k &=& \frac{AE}{l} =\frac{\frac{d^2 \pi}{4}\cdot 2.07\cdot 10^{11}}{20}\nonumber\\ k &=& \frac{\frac{\pi}{4}(0.01)^2 \cdot 2.07\cdot 10^{11}}{20} = 0.8129 \cdot 10^6\left[\frac{\mathrm{N}}{\mathrm{m}}\right]\nonumber \end{eqnarray} The natural freuqency is equal to: \begin{eqnarray} \omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{0.8129\cdot 10^6}{1000}} = 28.5114 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} The winch suddenly stopped wile it had velocity of \(2\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\). The period if ensuing vibration can be calcualted as: \begin{eqnarray} \tau_n &=& \frac{2\pi}{\omega_n} = \frac{2\pi}{28.5114} = 0.2204 \left[\mathrm{s}\right]\nonumber \end{eqnarray} The ampltiude is equal to: \begin{eqnarray} A &=& \frac{\dot{x}_0}{\omega_n} = \frac{2}{28.5114}\nonumber\\ A &=& 0.07015 \left[\mathrm{m}\right]\nonumber \end{eqnarray} Example 6 A bungee jumper (mass 100 kg) ties one end of an elastic rope of length 100 m and stiffnes 5 kN/m to a bridge and the other end to himself and jumps from the bridge . Assuming the bridge to be rigid, determine the vibratory motion of the jumper about his static equilibrium postion. \begin{eqnarray} G &=& 100 \cdot 9.81 = 981 \left[\mathrm{N}\right]\nonumber\\ l &=& 100 \left[\mathrm{m}\right] \nonumber\\ k &=& 5 \left[\mathrm{kN/m}\right]\nonumber\\ mgh &=& \frac{1}{2}mv^2 \nonumber\\ v &=& \sqrt{2gh} \nonumber\ v &=& \sqrt{2\cdot 9.81 \cdot 100} = 44.2945 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\nonumber\\ x_0 &=& x(t=0) = 0\nonumber\\ \dot{x}_0 &=& \dot{x}(t=0) = 44.2945 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\nonumber\\ x(t) &=& A_0 \sin(\omega_n t +\phi_0)\nonumber\\ A_0 &=& \sqrt{x_0^2 + \left(\frac{\dot{x}_0}{\omega_n}\right)^2} \nonumber\\ A_0 &=& \frac{\dot{x}_0\sqrt{m}}{\omega_n} \nonumber\\ A_0 &=& \frac{44.2945}{\sqrt{2000}}\sqrt{100} = 99.0455 \left[\mathrm{m}\right]\nonumber\\ \phi_0 &=& \tan^{-1} \end{eqnarray}
- the amplitude,
- the initial velocity,
- the maximum acceleration, and
- the phase angle
\begin{eqnarray} m\ddot{x} + kx &=& 0\nonumber \end{eqnarray} The general solution to previous differential equation can be written as: \begin{eqnarray} x &=& A\cos(\omega_n t - \phi) \end{eqnarray} a) The Amplitude The linear velocity is \(10 \left[\frac{\mathrm{cm}}{\mathrm{s}}\right] = 0.1 \left[\frac{\mathrm{m}}{s}\right]\), and the period of \(2 [\mathrm{s}]\). The amplitude can be found from relation \(v = \omega_n\cdot A \). However the \(\omega_n\) must be calculated from period. \begin{eqnarray} \tau_n &=& \frac{2\pi}{\omega_n}\Big/\cdot\frac{\omega_n}{\tau_n}\nonumber\\ \omega_n &=& \frac{2\pi}{\tau_n} = \frac{2\pi}{2} = 3.1416 \left[\frac{rad}{s}\right]\nonumber \end{eqnarray} Now the amplitude can be calculated: \begin{eqnarray} v &=& \omega_n\cdot A\Big/\cdot \frac{1}{\omega_n}\nonumber\\ A &=& \frac{v}{\omega_n} = \frac{0.1}{3.1416} = 0.03183 \left[\mathrm{m}\right].\nonumber \end{eqnarray} d) the phase angle The body with mass \(m\) is released at initial displacement of \(0.02\left[\mathrm{m}\right]\): \begin{eqnarray} x(t=0) &=& 0.02 \left[\mathrm{m}\right]\nonumber\\ x(0) &=& A\cos(-\phi_0) = 0.02\nonumber\\ \cos(-\phi_0) &=& \frac{0.02}{A} = 0.6283\nonumber\\ \phi_0 &=& 51.0724°\nonumber. \end{eqnarray} b) the initial velocity \begin{eqnarray} \dot{x}_0 &=& \dot{x}(t = 0) = -\omega_nA\sin(-\phi_0)\nonumber\\ \dot{x}-0 &=& -0.1 \sin(-51.0724°) = 0.07779 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\nonumber \end{eqnarray} c) the maximum acceleration - for maximum acceleration the general solution must be derived twice wit respect to time. \begin{eqnarray} \dot{x} &=& -A\omega_n \sin(\omega_n t+\phi_0)\nonumber\\ \ddot{x} &=& -A\omega_n^2 \cos(\omega_n t + \phi_0)\nonumber \end{eqnarray} To calculate the maximum acceleration the trigonometric part in previous equation is neglected nad absolute value of \(-\omega_n^2 A\) is used. Now using the second equation the acceleration is equal to: \begin{eqnarray} \ddot{x} &=& \omega^2 A = (3.1416)^2\cdot(0.03183) = 0.314151 \left[\frac{\mathrm{m}}{\mathrm{s}^2s}\right]\nonumber \end{eqnarray} Example 2 - an automobile having a mass of \(5000 \left[\mathrm{kg}\right]\) deflects its suspension springs \(0.05 \left[\mathrm{m}\right]\), under static conditions. Determine the natural frequency of the automobile in the vertical direction. Damping is negligible. \begin{eqnarray} \omega_n &=& \sqrt{\frac{g}{\delta_{st}}}\nonumber\\ \omega_n &=& \sqrt{\frac{9.81}{0.05}} = 196.2 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} Example 3 A rigid block of mass M is mounted on four elastic supports, as shown in Figure 2. As mass \(m\) drops from a height \(l\) and adheres to the rigid block without rebounding. If the sping of elastic support is \(k\), find the natural freuqnecy of vibration of the system:
- a) without the mass \(m\),
- b) with the mass \(m\).
In case b) with mass \(m\) the natural frequency can be written as: \begin{eqnarray} \omega_n &=& \sqrt{\frac{4k}{M+m}}.\nonumber \end{eqnarray} Velocity of the falling mass: \begin{eqnarray} v &=& \sqrt{2gl}\nonumber \end{eqnarray} At static equilibrium postion the \(x = 0\). At \(x_0 (t = 0) = -\frac{mg}{4k}\). Conservation of momentum: \begin{eqnarray} (M+m)\dot{x}_0 &=& mv = m\sqrt{2gl}\nonumber\\ \dot{x}_0 &=& \dot{x}(t=0) = \frac{m}{M+m} \sqrt{2gl}\nonumber \end{eqnarray} Complete solution: \begin{eqnarray} x(t) &=& A_0\sin(\omega_n t+\phi_0)\nonumber\\ A_0 &=& \sqrt{x_0^2 + \left(\frac{\dot{x}_0}{\omega_n}\right)^2}\nonumber\\ A_0 &=& \sqrt{\frac{m^2 g^2}{16k^2} + \frac{m^2 gl}{2k(M+m)}}\nonumber\\ \phi_0 &=& \tan^{-1}\left(\frac{x_0 \omega_n}{\dot{x}_0}\right) = \tan^{-1}\left(\frac{-\sqrt{g}}{\sqrt{glk(M+m)}}\right)\nonumber \end{eqnarray} Example 4 A hammer with mass \(6 \left[\mathrm{kg}\right]\) strikes anvil of mass \(50 \left[\mathrm{kg}\right]\) with a velocity of \(15\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\) as shown in Figure 3. The anvil is uspported on four springs, each of stiffness \(k = 17.5\left[\frac{\mathrm{kN}}{\mathrm{m}}\right]\). find the resulting motion of the anvil a) if the hammer remains in the contact with the anvil and b) if the hammer does not remain in contact with the anvil after the initial impact.
Solution: The velocity of the hammer is \(v = 15\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\), the \(x=0\) at static equilibrium position. \begin{eqnarray} x_0 &=& x(t=0) = -\frac{\mathrm{weight}}{k_{eq}} = -\frac{mg}{4k}\nonumber \end{eqnarray} Conservation of momentum: \begin{eqnarray} (M+m)\dot{x}_0 &=& mv\nonumber\\ \dot{x}_0 &=& \frac{mv}{M+m}\nonumber \end{eqnarray} Natural frequency: \begin{eqnarray} \omega_n &=& \frac{4k}{M+m}\nonumber \end{eqnarray} Complete solution can be written as: \begin{eqnarray} x(t) &=& A_0 \sin (\omega_n t + \phi_0)\nonumber \end{eqnarray} where: \begin{eqnarray} A_0 &=& \sqrt{x_0^2 + \left(\frac{\dot{x}_0}{\omega_n}\right)^2}\nonumber\\ &=& \sqrt{\frac{m^2 g^2}{16k^2} + \frac{m^2 v^2}{(M+m)4k}}\nonumber\\ \phi_0 &=& \tan^{-1}\left(-\frac{g\sqrt{M+m}}{v\sqrt{4k}}\right)\nonumber \end{eqnarray} \(v= 15\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\), \(m = 6 \left[\mathrm{kg}\right]\), \(M = 50 \left[\mathrm{kg}\right]\), \(k = 17.5 \left[\frac{\mathrm{kN}}{\mathrm{m}}\right]\).
\begin{eqnarray} A_0 &=& \sqrt{\frac{6^2 \cdot 9.81^2 }{16 \cdot 17.5^2} + \frac{6^2 \cdot 15^2 }{(50 + 6)\cdot 4 \cdot 17.5\cdot 10^3}}\nonumber\\ &=& 0.0454646 \left[\mathrm{m}\right]\nonumber\\ \phi_0 &=& \tan^{-1}\left(-\frac{9.81\cdot \sqrt{50+6}}{15\sqrt{4\cdot 17.5\cdot 10^3}}\right)\nonumber\\ &=& -0.0184958\nonumber \end{eqnarray} b) At static equilibrium postion: \(x_0 = x(t=0) = 0\) Conservation of momentum gives: \begin{eqnarray} M\dot{x}_0 &=& mv\nonumber\\ x_0 &=& x(t=0) = \frac{mv}{M}\nonumber\\ x_0 &=& x(t=0) = 0\nonumber \end{eqnarray} The amplitude: \begin{eqnarray} A_0 &=& \frac{mv}{\sqrt{4kM}} = \frac{6\cdot 15}{\sqrt{4\cdot 17.5\cdot 10^3 \cdot 50}}\nonumber\\ A_0 &=& 0.048107 \left[\mathrm{m}\right]\nonumber\\ \phi_0 &=& \tan^{-1}\left(\frac{x_0\omega_n}{\dot{x}_0}\right) = \tan^{-1}(0) = 0 \end{eqnarray} Example 5 A heavy machine weighing 9810 N is being lowered vertically down by a winch at velocity of 2 m/s. The steel cable supporting the machine has a diameter of 0.01 m. The winch is suddenly stopped when the steel cable's length is 20 m. Find the period of amplitude of the ensuing vibration of the machine. Solution: The mass of the machine can be calculated by dividing the machine weight by gravitational acceleration \(9.81\left[\frac{\mathrm{m}}{\mathrm{s}^2}\right]\) i.e: \begin{eqnarray} m &=& \frac{G}{g} =\frac{9810}{9.81} = 1000 \left[\mathrm{kg}\right]\nonumber \end{eqnarray}
To calculate the natural frequency first the stiffness must be calculated. The diameter of the cable is \(0.01 \left[\mathrm{m}\right]\), and the cable is made from steel so the modulus of elasticity is equal to \(2.07\cdot 10^{11} \left[\mathrm{Pa}\right]\) \begin{eqnarray} k &=& \frac{AE}{l} =\frac{\frac{d^2 \pi}{4}\cdot 2.07\cdot 10^{11}}{20}\nonumber\\ k &=& \frac{\frac{\pi}{4}(0.01)^2 \cdot 2.07\cdot 10^{11}}{20} = 0.8129 \cdot 10^6\left[\frac{\mathrm{N}}{\mathrm{m}}\right]\nonumber \end{eqnarray} The natural freuqency is equal to: \begin{eqnarray} \omega_n &=& \sqrt{\frac{k}{m}} = \sqrt{\frac{0.8129\cdot 10^6}{1000}} = 28.5114 \left[\frac{\mathrm{rad}}{\mathrm{s}}\right]\nonumber \end{eqnarray} The winch suddenly stopped wile it had velocity of \(2\left[\frac{\mathrm{m}}{\mathrm{s}}\right]\). The period if ensuing vibration can be calcualted as: \begin{eqnarray} \tau_n &=& \frac{2\pi}{\omega_n} = \frac{2\pi}{28.5114} = 0.2204 \left[\mathrm{s}\right]\nonumber \end{eqnarray} The ampltiude is equal to: \begin{eqnarray} A &=& \frac{\dot{x}_0}{\omega_n} = \frac{2}{28.5114}\nonumber\\ A &=& 0.07015 \left[\mathrm{m}\right]\nonumber \end{eqnarray} Example 6 A bungee jumper (mass 100 kg) ties one end of an elastic rope of length 100 m and stiffnes 5 kN/m to a bridge and the other end to himself and jumps from the bridge . Assuming the bridge to be rigid, determine the vibratory motion of the jumper about his static equilibrium postion. \begin{eqnarray} G &=& 100 \cdot 9.81 = 981 \left[\mathrm{N}\right]\nonumber\\ l &=& 100 \left[\mathrm{m}\right] \nonumber\\ k &=& 5 \left[\mathrm{kN/m}\right]\nonumber\\ mgh &=& \frac{1}{2}mv^2 \nonumber\\ v &=& \sqrt{2gh} \nonumber\ v &=& \sqrt{2\cdot 9.81 \cdot 100} = 44.2945 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\nonumber\\ x_0 &=& x(t=0) = 0\nonumber\\ \dot{x}_0 &=& \dot{x}(t=0) = 44.2945 \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\nonumber\\ x(t) &=& A_0 \sin(\omega_n t +\phi_0)\nonumber\\ A_0 &=& \sqrt{x_0^2 + \left(\frac{\dot{x}_0}{\omega_n}\right)^2} \nonumber\\ A_0 &=& \frac{\dot{x}_0\sqrt{m}}{\omega_n} \nonumber\\ A_0 &=& \frac{44.2945}{\sqrt{2000}}\sqrt{100} = 99.0455 \left[\mathrm{m}\right]\nonumber\\ \phi_0 &=& \tan^{-1} \end{eqnarray}