Does resonator will always dampen the driving oscillator? I was doing a question about resonance of sound box connecting to a tuning fork. It asks why the sound lasts for a shorter time with sound box than if the tuning fork is struck identically without the sound box.
Is it because the sound box dampens the vibration of tuning fork, or because due to larger contact area of box with air the box transfers vibrational energy at higher rate to air, or these two reasons are actually the same thing? So does it mean that resonator will always dampen the driving oscillator? I'm quite confused between the idea of energy transfer in resonance and damping. Thanks in advance:)
 A: This is like a driven harmonic oscillator problem where driver is the tuning fork. If the sound box have same frequency as of tuning fork the energy stored in tuning fork will quickly transfer into sound box, then it will dampen subsequently and that is the reason for faster damping. 
On the other hand if you provide energy to sound box then the sound box will resonate at its natural frequency and upon connecting tuning fork the sound box will act like driver. In this case the energy from sound box is transferred to tuning fork. 
It may be noted here that each resonating device has something called Q factor (or quality factor), which is decided by the dissipating term (first order term) in the harmonic oscillator problem. Larger the Q factor (smaller dissipation) sharper and stronger is the resonance. Hence if your sound box has very large Q factor then the energy is transferred from tuning fork to sound box super quickly. However the energy will start oscillating back and forth in fork and sound box (because damping is too low). The sound box will absorb only those frequencies which are very close to its natural frequency for other frequencies sound box will act as any other struck surface. 
If Q factor is small then energy transfer will be at lower rate but in larger band, however damping is high. 
Hope this will help
