Q7.10 Still more mass-spring systems (H)

The accompanying figures show three ways of using two springs to attach a mass m to supporting struts. [In case (C) the two springs are joined end-to-end]. The springs are identical, each having spring constant k. Find the angular SHM frequency of small oscillations of the mass, for each case
Graphic - No title - moresprings
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Hint

You need to think carefully about the extensions of the various springs when the mass is displaced by x. And you need to think about the directions of the forces they exert on the mass.
 

Solution

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Solution

These examples take you just one step beyond the oh-so-familiar mass spring problem of The familiar mass spring system. They are straightforward provided you apply the rules systematically.
  • [A] The equilibrium position is clear: the springs are unstretched. Let x represent a displacement of the mass, to the right from the this position.

    When the mass is displaced by x each spring exerts a restoring force of magnitude kx. The net force acting on the mass is then F=-2kx.

    Newton’s 2nd law then gives the equation of motion (EOM):

    \[ m\ddot{x} = -2kx \]
    which is of the SHM form
    \[ \ddot{x} = -\omega^2 x \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega=\sqrt{\frac{2k}{m}} \]

  • [B] The equilibrium position is clear: the mass is at the mid point so that the two springs are equally stretched. If the distance between the two walls is chosen appropriately, both springs can be in their unstretched state at this point. Let us assume that is the case (although we do not have to).

    Again let x represent a displacement of the mass, to the right from the position of equilibrium. We must think carefully about the forces exerted by each spring. As we do so we will think of x as a positive quantity (although the equations all work if x is negative, signifying a displacement to the left).

    When the mass is displaced to the right by x the left hand spring is stretched by x and pulls back with a force of magnitude kx. Thus the force it exerts along the positive x direction (to the right!) is

    FLH=-kx
    When the mass is displaced to the right by x the right hand spring is compressed by x and pushes back with a force of magnitude kx. Thus the force it exerts along the positive x direction (to the right!) is also
    FRH=-kx
    Newton’s 2nd law then gives the EOM:
    \[ m\ddot{x} = -2kx \]
    which is again of the SHM form
    \[ \ddot{x} = -\omega^2 x \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega=\sqrt{\frac{2k}{m}} \]
    The oscillation frequency is the same as in case (A).

  • [C] In equilibrium both springs are unstretched. Again let x represent a displacement of the mass, to the right from the position of equilibrium. Then each spring is stretched by x/2. The only force acting on the mass is that exerted by the spring with which it is in contact. [If you think that the other spring has to be included here too you haven’t yet absorbed the rules for applying the 2nd Law –in particular the idea of drawing a box to define your ‘system’: review S2.3.] This force is restoring and has magnitude kx/2 The net force acting on the mass is thus

    \[ F=-k\frac{x}{2} \]
    and the EOM is
    \[ m\ddot{x} = -k\frac{x}{2} \]
    which is again of the SHM form
    \[ \ddot{x} = -\omega^2 x \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega=\sqrt{\frac{k}{2m}} \]

    Note that two springs acting ’in series’ like this (case C) give a lower frequency than a single spring, while two springs arranged ’in parallel’ (case A) give a higher frequency. You might like to think how these results generalise to the case of n springs?

    One other remark. Some students think that they can answer case C simply by saying that the two springs, joined up this way, are just ’like’ a single, but longer spring. So the result should simply be the familar one for the single-spring case. This is true but misleading; the two springs do indeed ‘act’ like a single spring; but the spring constant for that single spring is not the same as that for the two springs out of which it is composed. A spring constant reflects not just the material but also the geometry of the spring. Our argument shows that if you double the length of a spring by joining it up to another identical one the spring constant of the long one is half that of the two springs from which it is built.