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General Physics I:

Section 7.6 Examples C

Example 7.6.1.

A mass of 1 kg resting on a frictionless horizontal surface is pressed against a spring of constant 100 N/m compressing the spring 25 cm and released from rest as shown in the figure below.
  1. Where will the block attain its fastest velocity?
  2. How fast will it be moving at that time?
  3. Where will it stop again?
  4. What will it do after it stops?
  5. Where does the block attain its greatest acceleration?
  6. What is that acceleration?
  7. Describe the motion of the block as time progresses.
  8. Write the equation of motion for the block.
  9. Write the equation for the velocity of the block.
  10. Write the equation for the acceleration of the block.
  11. Compare (h) with (j) to obtain a relationship between \(a\) and \(x\) for the block.
  12. Apply Newton’s Second Law to the block so as to find \(a\) in terms of \(x.\)
  13. Compare (k) with (l) to obtain a relationship between \(k\) and \(m\) at (\(x=A\)) for the block.
  14. How long does it take for the block to reach the equilibrium point the first time?
  15. How fast is the block moving when it is 10 cm beyond the equilibrium point?
  16. What is the energy of the block when it is 10 cm beyond the equilibrium point?
Solution.
Given: \(m = 1 \,kg, \quad k = 100 \,N/m,\quad A = 0.25 \,m,\quad v_{o} = 0,\) \(\omega = \sqrt{\frac{k}{m}}=10 \,rad/s. \)
  1. The velocity particle in shm is given by
    \begin{equation*} v=A\omega\sin\omega t. \end{equation*}
    For \(v\) to be maximum, \(\sin\omega t=1 \Rightarrow\omega t=\frac{\pi}{2}\text{.}\) Hence the position of the particle at that time is
    \begin{equation*} x=A\cos\omega t = A\cos\frac{\pi}{2}=0 \end{equation*}
    i.e., at the equilibrium position: \(x = 0.\)
  2. At equilibrium position, velocity of the block is \(v=A\omega = 0.25\times10= 2.5 \,m/s.\)
    Alternatively: The total potential energy at extreme position is converted into total kinetic energy at mean position, i.e.,
    \begin{equation*} 1/2 k A^{2} = 1/2 m v^{2} \quad \Rightarrow \quad v = \omega A = 2.5 \,m/s \end{equation*}
  3. Velocity,
    \begin{equation*} v=A\omega\sin\omega t =0 \quad \Rightarrow \quad \omega t=0. \end{equation*}
    Therefore,
    \begin{equation*} x=A\cos\omega t = A. \end{equation*}
    Hence at 25 cm to the right of equilibrium: \(x = 0.25 \,m.\)
  4. Because of maximum restoring force it starts accelerate to the left.
  5. Since the restoring force is maximum at extremes, it attains maximum acceleration (\(a=-\omega^{2}x\)) at that points. Hence at the extremes, where it stops: \(x = A= \pm 0.25 \,m.\)
  6. \begin{equation*} k x = m a \Rightarrow a = \omega^{2}A = 25 \,m/s^{2} \end{equation*}
  7. It speeds up, coasts through equilibrium, slows down, stops, and reverses itself.
  8. \begin{equation*} x = - A \cos\omega t, \quad \omega = \Rightarrow x = - (0.25 \,m) \cos((10 \,rad/s) t) \end{equation*}
  9. \begin{equation*} v = dx/dt = \omega A \sin\omega t \Rightarrow v = (2.5 \,m/s) \sin ((10 \,rad/s) t) \end{equation*}
  10. \begin{equation*} a = \,d^{2}x/\,dt^{2} = \omega^{2} A \cos\omega t \Rightarrow a = (25 \,m/s^{2}) \cos ((10 \,rad/s) t) \end{equation*}
  11. \begin{equation*} \frac{a}{x} = -100 = -\omega^{2} \quad \Rightarrow a=-\omega^{2} x \end{equation*}
  12. \begin{equation*} - k x = m a \Rightarrow a = - \frac{k}{m} A \end{equation*}
  13. \begin{equation*} \frac{a}{a} = \frac{-\omega^{2} A}{- \frac{k}{m} A}\quad \Rightarrow\quad \omega^{2} = k/m \quad \Rightarrow \omega = \sqrt{\frac{k}{m}} \end{equation*}
  14. \begin{equation*} [x = - (0.25 m) \cos ((10 rad/s) t) = 0 \Rightarrow \cos 10t =\cos \frac{\pi}{2} \end{equation*}
    \begin{equation*} \quad \therefore t=\frac{\pi}{20} = 0.157 \,s \end{equation*}
  15. \begin{equation*} 1/2 k A^{2} = 1/2 m v^{2} + 1/2 k x^{2}, \quad \text{at}\quad x = 0.1 \,m, \end{equation*}
    \begin{equation*} v = \omega\sqrt{A^{2}-x^{2}}= 2.29 \,m/s \end{equation*}
  16. \begin{equation*} E = \frac{1}{2} k A^{2} = 3.125 \,J \end{equation*}

Example 7.6.2.

A pendulum is hanging from the roof of a tall building which oscillates 10 times in 1 minute, find the height of the building.
Solution.
Given: \(T= 1 minute= 60 s, \quad g=9.8m/s^{2},\)
\begin{equation*} f=10 cycle/minute= \frac{10}{60} (cycles/sec.) = \frac{1}{6} \,Hz. \end{equation*}
\begin{equation*} \omega = 2\pi f = 2\pi\frac{1}{6}=\frac{\pi}{3} = 1.05 \,rad/s \end{equation*}
\begin{equation*} \therefore \quad \omega =\sqrt{\frac{g}{l}} \end{equation*}
\begin{equation*} or, \, l =\frac{g}{\omega^{2}} =\frac{9.8 }{1.05^{2}} = 8.89 \,m \end{equation*}

Example 7.6.3.

A spring-mass system executing SHM with a period of 0.80 s and has a maximum displacement of 10 cm. Initially the mass is at 5.0 cm left of equilibrium and moving to the left.
  1. What is its position at \(t=2.0 \,s\) and in what direction it is moving at that time?
  2. What is its velocity and direction at that time?
Solution.
Initial condition at \(t=0,\)
\begin{equation*} x_{o}=-5.0 cm = A\cos\phi_{o} \end{equation*}
\begin{equation*} \therefore \quad \phi_{o}=\cos^{-1}\left(\frac{x_{o}}{A} \right) \end{equation*}
\begin{equation*} = \cos^{-1}\left(-\frac{1}{2} \right) =\pm\frac{2\pi}{3} \,rad \end{equation*}
Hence,
\begin{equation*} \phi_{o} =\frac{2\pi}{3} \,rad \end{equation*}
\begin{equation*} \omega=\frac{2\pi}{T} = \frac{2\pi}{0.80} = 7.85 \,rad/s \end{equation*}
  1. \begin{equation*} x(t) = A\cos\left(\omega t +\phi_{o}\right) = 10 \cos\left(7.85\times 2 +\frac{2\pi}{3}\right) = 5.0 \,s \end{equation*}
    to the right of mean position.
  2. \begin{equation*} v(t) =-\omega A\sin(\omega t +\phi_{o}) \end{equation*}
    \begin{equation*} = -7.85\times 10 \sin\left(7.85\times 2 +\frac{2\pi}{3}\right) = 68 \,cm/s \end{equation*}
    to the right of mean position going to towards \(+A.\)