# What is the effect of mass on resonance amplitude?

When a system is undergoing forced oscillations, why does reducing the mass of the system cause the frequency response curve to shift downwards? I encountered this problem in a practice paper, but I can't seem to put my head around the model answer. I get that reducing the mass causes the natural frequency to increase, so the resonance frequency becomes higher, but why does the amplitude of oscillation decrease across all frequencies?   • See en.wikipedia.org/wiki/… and the solution for driven oscillators. There you see that the amplitude is damped by an exponential decrasing which is stronger if the mass is smaller. – Alpha001 Aug 11 '18 at 13:12
• But what if the system is not damped? – Anonymous Aug 11 '18 at 13:19
• Look at cts.iitkgp.ac.in/home/pratik/forcednew.pdf for it. Then the amplitude $A = \frac{F}{m(\omega_0^2-\omega^2)}$ depends also on the mass $m$ and the force $F$. – Alpha001 Aug 11 '18 at 13:38
• "But what if the system is not damped?" Take the limit of the damped behavior as $b \to 0$ and make your math teacher cry. Real systems, of course always have some damping and will break if you make the driving force sufficiently large in comparison to the damping forces. – dmckee Aug 11 '18 at 15:57

## 1 Answer

This system does have losses - if it did not, the amplitude at resonance would be increasing until it hit the limits, i.e., the springs would have no further way to compress.

In a steady state, all power delivered by the motor is spent to cover for these losses.

If the amplitude of the oscillations at higher frequencies was the same as at the amplitude at lower frequencies, the distance traveled by the trolley and all other parts of the system, per unit time, would be greater at higher frequencies and, therefore, assuming that the losses are proportional to the distance, more power would have to be spent.

Since the power of the motor in (iv)(2) is maintained, while the frequency is increased, we have to conclude that the amplitude has to decrease.