If we consider an object undergoing who has an acceleration proportional to the displacement of the object, it is going simple harmonic motion.

In terms of Newton's second law, this is $$ -\dfrac k mx = \dfrac{d^2 x}{dt^{2}} $$ where $k/m := \omega^2$, the frequency.

For an object floating because of the buoyant forces, Netwton's second law looks like $F_B - F_g = ma$. Why can we consider this SHM as they do here (for example: SHM of floating objects)?

  • $\begingroup$ In fact, that should be $kx = \dfrac{d^2 x}{dt^{2}} \cdot m$ $\endgroup$
    – Inceptio
    May 11, 2013 at 5:23
  • $\begingroup$ @Inceptio I was kind of considering $k$ to include the $m$ but I'll update the question for clarity. $\endgroup$
    – jaynp
    May 11, 2013 at 7:27

1 Answer 1


Consider an object that is floating and stationary in a liquid of density $\rho$ such that a volume $V_s$ of the objects is submerged. Suppose that we orient a cartesian coordinate system such that the positive $z$-axis points upward, orthogonal to the surface of the liquid.

When the object is floating at rest, the net force it experiences is zero. If we displace the object by a small amount $z$ from its equilibrium position, then the submerged volume will change by an amount $$ \delta V_s = -A z $$ where $A$ is the cross-sectional area of the portion of the object intersecting the surface of the water. The buoyant force experienced by the displaced object will have changed by an amount equal to the change in the weight of the displaced liquid $$ \delta F_B = \rho\delta V_s g = -\rho gA z $$ Since the net force on the object is the buoyant force minus the force due to gravity, and since it is only the buoyant force that changes for the displaced object, the net force on the object when it is displaced by an amount $z$ is $$ F(z) = -\rho gA z $$ Applying Newton's second law in the $z$-direction gives $$ -\rho gA z = m\ddot z $$ which is precisely the equation for simple harmonic motion.

  • $\begingroup$ How would this derivation change, if the downward direction is taken to be positive $z$ ? $\endgroup$
    – RayPalmer
    Jan 5, 2022 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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