Q7.17 Consolidation exercise (K)

A 2 kg mass is attached to a spring and placed on a smooth horizontal surface. A horizontal force of 20 N is required to hold the mass at rest when it is pulled 0.2 m from its equilibrium position. The mass is now released from this position and it undergoes simple harmonic oscillations. Find
  1. The spring constant of the spring.
  2. The frequency of oscillations.
  3. The maximum speed of the mass (where does it occur?).
  4. The maximum acceleration (where does it occur?).
  5. The total energy.
  6. The velocity and acceleration when the displacement is one-third of the amplitude.
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Hint

This is a further tour of the basics. Try to do it without appeal to your notes.
 

Solution

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Solution

Now, just once more with feeling...
  1. When displaced to x=+0.2m the spring exerts a force of -20N where the minus sign implies that the force acts in the opposite direction to the displacement. The spring constant is then
    \[ k = \frac{-F}{x} = \frac{20 N}{0.2m} = 100 Nm^{-1} \]
  2. From Key Point 6.4 the frequency of the resulting oscillations is
    \[ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{100 Nm^{-1}}{2kg}} = 1.13Hz \]
  3. From the general SHM solution (Key Point 6.3)
    x=xmcos(ωt+φ)
    the velocity
    \[ v=\dot{x} = -\omega x_m\sin(\omega t + \phi) \]
    so the speed has a maximum value
    vmax=xmω=(0.2m)(2π)(1.13Hz)=1.41ms-1
    It has this value whenever the mass passes through the equilibrium position.
  4. The acceleration follows either by differentiating the velocity or by direct appeal to the generic SHM equation (Key Point 6.2)

    \[ \ddot{x} = -\omega ^2 x \]
    Its maximum value is
    amax=ω2xm=(2πf)2xm=4π2(1.13 Hz)2(0.2 m)=10.1 ms-2

    The acceleration has this magnitude whenever the particle is at an extreme position.

  5. Since the particle is released from rest the total energy is equal to the initial PE:
    \[ E = \frac{1}{2}kx_m^2 = \frac{1}{2}(100Nm^{-1})(0.2m)^2 = 2\,J \]
  6. When x=xm/3 the PE is
    \[ U(x=x_m/3) = \frac{1}{2}kx_m^2 \times \frac{1}{9} = \frac{E}{9} \]
    At this point then
    \[ \frac{1}{2}mv^2 = K = \frac{8E}{9} \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} v= \pm \sqrt{\frac{16E}{9m}}= \pm 1.33 ms^{-1} \]
    Finally, the acceleration at x=xm/3 is
    \[ a= \ddot{x} = -\omega^2 x = -\omega^2 \frac{x_m}{3} = - 3.36\,ms^{-2} \]