3
$\begingroup$

If a block of mass $M$ is attached to a spring of mass $m$ and its force constant is $k$, if the system oscillates horizontally on a frictionless ground. We could derive the period of the oscillation as $$T=2\pi\left(\frac{M+m/3}{k}\right)^{1/2}$$

What I am confused is if we draw a FBD of the block of mass, we would get this equation according Newton's second law $$-kx=Ma$$ while $x$ equals to the displacement from the equilibrium and $a$ equals to the acceleration of the mass block.

Which is the same equation when the spring does not have mass.

As a result, the period of the block of mass would be $$\tau = 2\pi \left(\frac{M}{k}\right)^{1/2}$$ which is not the same as the period $T$.

Please explain why the results is not the same. Thank you very much.

$\endgroup$
6
  • 1
    $\begingroup$ Did your force diagram show (i) the spring's weight & (ii) the restoring force on each part of the spring from those parts pulling on it? (Hint: the reason a $\tfrac13$ factor appears is because it's $\int_0^1u^2du$.) $\endgroup$
    – J.G.
    Nov 16 '21 at 8:09
  • $\begingroup$ If we draw a force diagram of a block of mass do we have to consider the spring's weight? I thought the force is simple $-kx$ due to the stretched distance of the spring. $\endgroup$
    – RKJ
    Nov 16 '21 at 9:19
  • 1
    $\begingroup$ You have to show all forces on the system, yes. A differential equation in $x$ will emerge from every factor that affects how $x$ evolves. Would you be interested in alternative solution techniques, such as Lagrangian mechanics? $\endgroup$
    – J.G.
    Nov 16 '21 at 9:26
  • $\begingroup$ I have solved the correct period by using the conservation of energy (differertiate kinetic and potential energy respect to time equals 0), but what I am not understand is why does the free body diagram on the block of mass does not work, does that mean there are some other forces done on the block (Hooke's Law incompatible in this case)? $\endgroup$
    – RKJ
    Nov 16 '21 at 9:37
  • 1
    $\begingroup$ Imagine one end of the spring is fixed to a wall, the other is attached to the mass. The parts of the spring that are close to the wall provide a force that accelerates the block and the remaining part of the spring. The massless analysis does not take that into account. $\endgroup$
    – garyp
    Nov 16 '21 at 15:27
2
$\begingroup$

Hooke’s Law $F=kx$, where $F$ is the restoring force, $k$ Is the spring constant, and $x$ is the displacement, is defined for static equilibrium. For the special case of a massless spring, though, the relation can be used in dynamic situations as well and can therefore be applied to the analysis of simple harmonic motion, for example. The reason is that the massless spring responds instantaneously.

If we try to apply static Hooke’s Law to the dynamics of a massive spring, however, we run into problems because fast compression or elongation encounters not only the restoring force from spring elasticity but also an inertial force from needing to accelerate some of the spring mass (effectively one-third of the spring mass).

$\endgroup$
2
  • $\begingroup$ Your answer is really helpful, thank you very much. But I still have a little confusion, due to what you have said, if a massive spring is accelerated, the applied force done on the block of mass at one end can not express with $-kx$ because of the internal force in spring right? If that so, why does the potential energy of spring $1/2kx^2$ could still be used when considering the derivative of total energy (K.E.+P.E.) with respect to time equals 0? $\endgroup$
    – RKJ
    Nov 16 '21 at 16:28
  • $\begingroup$ Sorry—what else would the potential energy be? $\endgroup$ Nov 16 '21 at 20:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.