Why doesn't amplitude increase when the frequency of external periodic force increases above the natural frequency of the vibrating object?
Answer
In one sense this is a consequence of the meaning of resonance. Resonance is identified by a maximum in amplitude. If amplitude increased above (or below) the resonant frequency then we have not found a maximum in amplitude.
Perhaps what you are really asking is, why is there a maximum in amplitude when driving frequency $f$ equals the natural frequency $f_0$? This was answered by What is the qualitative cause for a driven oscillator to have a max. amplitude during resonance?
At the simplest level this is because at the natural frequency $(f=f_0)$ the driving force is always in phase with the natural motion of the system. The driving force is then always adding energy to the system, which will increase indefinitely unless there is some form of damping (eg friction) which removes energy from the system at a faster rate as amplitude increases.
At all other frequencies the driving force is sometimes in phase and sometimes out of phase with the natural motion of the system. Sometimes it adds energy, sometimes it takes energy away.
However, this assumes that both the natural motion and the driving force are both sinusoidal. If the driving force is applied as an intermittent impulse or "kick" then sub-harmonic frequencies of $f_0/n$, where $n$ is an integer, will also result in resonance. For example, if the system oscillates once every second, giving it a "kick" in the right direction once every 2 or 3 or 4 etc seconds will also increase the amplitude of the system, provided that the energy imparted with each kick is not outweighed by the loss of energy in between. See Non-resonant but efficient frequencies.
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