The amplitude of a lightly damped oscillator decreases by 3.0 percent during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?
We’re told that the oscillator’s amplitude decreases by 3.0 percent over each cycle of oscillation; we’ll call that value Δ𝐴. We want to know the percentage of mechanical energy lost in each cycle given this loss of amplitude; we’ll call that Δ𝐸.
We can start by sketching out what the amplitude of this oscillator might look like over time. If we sketch out the amplitude of this oscillator as a function of time, then we can see that as the oscillator goes up and down, over time the amplitude of this oscillations decreases slightly with each cycle.
That gradual constant decrease in amplitude over time is quantified by Δ𝐴, which tells us that over each complete cycle of the oscillator the amplitude lost is 3.0 percent.
To figure out Δ𝐸, the percentage change in mechanical energy of the oscillator over each cycle, let’s recall that for an oscillating system the power, 𝑃, of that oscillator is proportional to the amplitude of oscillation squared.
If we also recall that power is equal to energy used over time, then we can see it’s also true that the oscillators energy 𝐸 is proportional to 𝐴 squared as well. Applying this relationship to our scenario, the percentage change in mechanical energy of the oscillator over each cycle is proportional to the change in amplitude over each cycle as a percent squared.
So if, as in our case, Δ𝐴 is equal to 3.0 percent, then the oscillator’s mechanical energy will change as the square of that. So Δ𝐸, the change in mechanical energy of the oscillator over each cycle, is 3.0 percent squared or 9.0 percent. That’s how much mechanical energy the oscillator loses each cycle as a percentage of its total energy.