Impact of mass on resonance

From the equation of motion for driven, damped harmonic oscillator results that the amplitude is givn by $$ A=frac{F_m }{sqrt{m^2(omega^2-omega_d^2)^2 +b^2omega_d ^2 }} $$ where Fm is the aplitude of the driving force, $omega_d$ is the driving frequency, $omega$ is the oscillator's frequency and b is the damping factor. From this we can see that the amplitude of the resonance peak (at $omega_d$=$omega$) is $$ A_{max}=frac{F_m}{bomega_d}$$ which is independent of mass. However, the width of the resonance peak depends on the mass. If we take the half width at half peak, we get $$|omega_d^2-omega^2|= sqrt{3}frac{b omega_d}{m} $$ So larger mass means narower resonance peak. However, the context of your question may be different than the forced damped harmonic oscilaltor so it would be good if you would give more details in your post.

So in the non-driven damped harmonic oscillator:

$$ ddot x + 2gamma dot x + omega_0^2 x = 0 $$

where the undamped resonant frequency is:

$$ omega_0 = sqrt{frac k m } $$

and

$$ gamma = frac c {2sqrt{km}}$$

is tha dampin ratio ($c$ is the scale between damping force and velocity).

First of all, we see $gamma$ decreases with increasing mass. This makes sense, as the damping force is fixed by velocity, so if $m$ is larger, acceleration at fixed force is less.

The width of a resonance is usually quantified by the Q-factor:

$$ Q = frac 1 {2gamma} = frac{sqrt km} c $$

which increases with mass. Note that:

$$ Q = 2pi times frac{rm energy stored per cycle}{rm energy lost per cycle}$$

and the energy lost depends only on $dot x$, while the energy stored depends on $frac 1 2 x^2_{max}$, which depends on $frac 1 2 m dot x_{max}^2$.

That last part explains why a fixed velocity, the amplitude increases with mass: there is more kinetic energy, so the spring compresses/extends more (per the question).

But, claiming fixed velocity is not exactly fair. If you look at the driven oscillator with sinusoidal driving force (amplitude $F_0$, frequency $omega$), then amplitude is:

$$ x_{max} = frac F {momegasqrt{(2omega_0gamma)^2+frac{omega_0^2-omega^2}{omega^2} }}$$

which takes the value, on resonance:

$$ x_{max} = frac F {momega_0 (2omega_0gamma)} = frac F {2 momega_0^2 gamma} = frac F {2mfrac k m frac c {2sqrt{km}}} = frac F c sqrt{frac m k } = Q frac F k$$

which indeed increases with $sqrt m$. I would phrase this as the amplitude is proportional to the driving force divided by the spring constant (which makes perfect sense), and is then scaled by they Q-factor.