Q7.7 Consolidation exercise (K)

One end of a spring is fastened to a rigid support. A calibrated spring balance attached to the free end is used to determine that the restoring force is proportional to displacement with a force of 4 N giving a displacement of 0.02 m. A 2 kg object is then attached to the end of the spring, allowed to come to equilibrium, then pulled a distance of 0.04 m, and released.
  1. Find the spring constant k.
  2. Find the angular frequency, the frequency and the period of vibration.
  3. Find the maximum velocity and acceleration.
  4. How much time is required for the object to move halfway back to the centre from its initial position?
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Hint

You will find all you need in section S6.3.
 

Solution

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Solution

  1. The spring constant k prescribes the restoring force exerted by the spring, per unit displacement. The data provided shows that k×0.02=4 implying k=200 Nm-1
  2. Writng Newton’s second law for the mass, moving under the action of the spring we have
    ma=-kx
    implying
    \[ \ddot{x} = -\omega^2 x \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega = \sqrt{\frac{k}{m}} \]
    The value of the angular frequency is thus $\omega = \sqrt{\frac{200}{2}} = 10\,rad \cdot s^{-1}$ The period follows as
    \[ T = \frac{2\pi}{\omega}= 0.628s \]
    while the frequency is
    \[ f=\frac{1}{T} = 1.59\,Hz \]
  3. The general equation for the displacement (remember it!)
    x=xmcos(ωt+φ)
    implies the equation for the velocity (which, therefore, you don’t need to remember...as long as you can differentiate confidently!):
    \[ v=\dot{x}=-\omega x_m\sin(\omega t +\phi) \]
    which has its maximum when the sine function is unity. The maximum speed is thus
    vmax=ωxm=0.4ms-1
    Similarly, the acceleration
    \[ a = \ddot{x} = -\omega^2 x \]
    has maximum value
    amax=ω2xm=4ms-2
  4. The initial conditions are x(t=0)=xm and v(t=0)=0, since the particle is released from rest. The specific form of the general solution we need to match these initial conditions is
    x=xmcos(ωt)
    The time required to reach the point x=xm/2 is the solution of
    \[ \frac{x_m}{2} = x_m \cos (\omega t) \]
    implying
    \[ \omega t = \cos^{-1} \left(\frac{1}{2} \right) = \frac{\pi}{3} \]
    so t=π/30 s=T/6. Contrast this with the time it takes to reach the equilibrium point x=0, which is T/4. So it takes T/6 to travel from x=xm to x=xm/2 but only a further T/12 to travel from x=xm/2 to x=0. The distances covered in the two intervals is the same (namely xm/2); but the second one takes less time because the average speed is bigger.