Coil Resonant Frequency
The answer lies in the Fourier spectrum of the pulse and the self capacitance of the windings creating a bandpass filter or "tank resonator".

You ought to know a square wave consists of only odd harmonics.  Imagine that the 2f sine fits in the pulse width and thus cancels out and all even harmonics are also null.

When you reduce the pulse to an impulse you ought to know the spectrum is full spectrum limited by the risetime.

In between if you imagine the sine wave that fits in the pulse width, it will also be a null and harmonics of that f.  Below you see the cosine x /x  response of the envelope of harmonics on the bottom in LOG SCALE and the phase response of each harmonic as well.  

Example

Below   imagine I simulated a 1Hz narrow Pulse ( by mouse drawing on this website)  and it must be 1/20th of 1 Hz or 50ms because the 20the harmonic is a null. ( or close to it at -50dB)

Now to answer your question, when you change the pulse width you are injecting varying energy at the harmonics of your BPF so the amplitude goes up and down as the nulls sweep across the BPF centre frequency.

You can  go to the site in the image and draw any arbitrary repeating wave with your mouse or choose default types to recognize and remember the Fourier response of each pattern.  Note the triangle is the integral of the Square wave so the triangle spectrum indeed looks like a low pass filter applied to the square wave spectrum.  First select Mag/Phase then Log for Bode plot like images like a spectrum analyzer in dB and view any result with the mouse cursor.  YOu can even listen to it.