# How exactly can we shake an object to determine its mass?

It's often said that inertial mass is a measure of how hard it is to shake an object (as distinct from gravitational mass, which is how hard it is to lift an object in a gravitational field). Because this is so often said, one would think that at some point in my physics career I would have been asked to do a toy problem where I derive a formula for the inertial mass of an object based on how hard it is to shake. But I've never been assigned such a problem, so I sat down to try to perform the calculation myself.

It seems to me that "how hard it is to shake an object" could be made precise as the average power needed to induce oscillations at a certain amplitude with a certain period. So suppose we shake a mass so that its motion is given by $$x(t) = A\sin \left(\frac{2\pi t}{T}\right)$$ This implies an applied force $$F(t) = m\ddot{x}(t) = -\frac{4\pi^2 mA}{T^2}\sin \left(\frac{2\pi t}{T}\right)$$ And thus an instantaneous power $$P(t) = F(t)\dot{x}(t) = -\frac{8\pi^3 mA^2}{T^3}\sin \left(\frac{2\pi t}{T}\right) \cos \left(\frac{2\pi t}{T}\right)$$ which can be simplified to $$P(t) = -\frac{4\pi^3 mA^2}{T^3}\sin \left(\frac{4\pi t}{T}\right)$$ Averaging over a period, we obtain \begin{align} \langle P \rangle &= \frac{1}{T} \int_0^T-\frac{4\pi^3 mA^2}{T^3}\sin \left(\frac{4\pi t}{T}\right) \, dt \\&= -\frac{4\pi^3 mA^2}{T^4} \int_0^T \sin \left(\frac{4\pi t}{T}\right) \, dt \\&= \frac{\pi^2 mA^2}{T^3} \left[ \cos\left(\frac{4\pi t}{T}\right) \right]_0^T \\&= 0 \tag{!} \end{align} Disturbingly, we find that the average power is $$0$$, which makes it useless for calculating mass. This makes some sense, however, since we alternately do work with and against the motion of the mass.

I seem to recall from my time studying E&M that the thing to do in such situations is to instead calculate the RMS of the power, which yields \begin{align} P_{\text{RMS}} &= \frac{4\pi^3 mA^2}{T^3} \left[ \frac{1}{T} \int_0^T \sin^2 \left(\frac{4\pi t}{T}\right) \right]^{1/2} \\&= \frac{4\pi^3 mA^2}{T^3}\left[ \frac{1}{T} \cdot \frac{T}{2} \right]^{1/2} \\&= \frac{2\sqrt{2}\pi^3 mA^2}{T^3} \end{align} In terms of frequency $$\omega$$, we have $$P_{\text{RMS}} = \frac{mA^2\omega^3}{2\sqrt{2}}$$ which seems like a pretty reasonable equation. But unlike the E&M case where we are very naturally led to consider RMS quantities (since power goes like $$I^2$$), the introduction of RMS here seems sort of artificial and like a trick to avoid an unpleasant $$0$$.

In essence, my question is (1) does this RMS equation for power make precise the notion that inertial mass is how hard it is to shake something, and (2) is there a physical motivation for looking at RMS power in this situation?

• I don't think we have a good internal gauge for power, RMS or not (except maybe in cases of extended periods of exercise for example). I would think that the absolute value of force would be a better indicator. Of course the average or peak value of this is also proportional to mass, but it is different from (RMS) power in that it has a different dependence on $A$ and $\omega$.
– Puk
May 30, 2022 at 4:05
• @Puk maybe not an internal gauge, but is RMS mechanical power something easily measured? I suppose you could drive the mass with an AC motor, and then RMS power is a very natural thing to measure. May 30, 2022 at 4:14
• I suppose another thing to say is that simple dimensional analysis tells you that no matter how you look at power, you're going to get $P \sim mA^2 \omega^3$. May 30, 2022 at 4:18
• I think force might still be somewhat easier to determine electrically depending on the type of motor, because there will be various power losses (electrical and mechanical) in the motor.
– Puk
May 30, 2022 at 4:32