How does the time-energy uncertainty principle give particles a non-well-defined mass?

Question

It is example 3.7 from Griffith Quantum Mechanics 3ed:

The $\Delta$ particle last about $10^{-23}\ \mathrm{s}$, before spontaneously disintegrating. If you make a histogram of all measurements of its mass, you get a kind of bell-shaped curve centered at $1232\ \mathrm{MeV}/c^{2}$, with a width of about $120\ \mathrm{MeV}/c^{2}$ (figure: 3.2). Why does the rest energy ($mc^{2}$) sometimes come out higher than $1232$, and sometimes lower? Is this experimental error?

Ans: No, for if we take $\Delta t$ to be the lifetime of the particle (certainly one measure of "how long it takes the system to change appreciably"), $$\Delta E\Delta t = \left(\frac{120}{2}\ \mathrm{MeV}\right) \left(10^{-23}\right) = 6 \times 10^{-22}\ \text{MeV s}.$$ whereas $\frac{\hbar}{2}$ = $3 \times10^{-22}\ \text{MeV s}$. So the spread in $m$ is about as small as the uncertainty principle allows - a particle with so short a lifetime just doesn't have a very well-defined mass.

Now my question is what does Griffiths mean by: "So the spread in $m$ is about as small as the uncertainty principle allows"; and also why does he say: "a particle with so short a lifetime just doesn't have a very well-defined mass." What is the reason or logic behind this?

I would like an explanation from a physical point of view rather than a mathematical one.

Answer 1

E and t aren't operators, but you can still derive ΔEΔt ≥ℏ/2. So if a particle has a very short lifetime then ΔE must be big enough to satisfy the uncertainty principle and since E∝m the mass won't be well defined. -- This answer has given by (from comment): https://physics.stackexchange.com/users/284754/wihtedeka