Spring-mass system in an elevator I have a conceptual doubt.
If we have a mass m hanging from an ideal spring, attached to the ceiling of an elevator... what happens to the period if the elevator starts going up or down with constant acceleration?
I saw that the period did not change in any of these cases, but I don't know why. I imagined that due to an apparent force acting in this system, the frequency and period would change...
How to understand this?
In the case of a pendulum, the period would change, right?
 A: The force of a spring is directly proportional to the extension of the spring regardless of what is attached or other boundary conditions.
Force of the spring is always $F = F_0 + k x$ with preload $F_0$ and spring rate $k$.
The oscillation happens around the balance point of the spring so the preload does not matter. It is the spring rate of the spring and the attached mass it moves that cause it to behave with the following law
$$ \ddot{x} = - \left( \tfrac{k}{m} \right) \; x $$
Anything with $\ddot{x} = -\omega^2\, x$ will undergo SHM with frequency $\omega$ as a solution to the above equation.
Any acceleration of the base, only changes the balance point, but not what happens as a result of movement near the balance point.
A: You brought up two situations: spring-mass and pendulum.
For the pendulum: Yes the frequency changes.
They have different answers. The pendulum frequency will change. The spring-mass frequency will not.
Spring-Mass
For a spring mass, if accelerating up at $a=\frac{d^2z}{dt^2}$, where $a$ can be positive, or negative (accelerating down), the entire oscillation trajectory will be displaced by:
$$\Delta z=-\frac{F}{k}= \frac{-ma}{k}$$
That applies to the highest point, the lowest point, the midpoint - the entire motion will just be displaced $\frac{-ma}{k}$ in the Z direction, opposite to the elevator’s direction of acceleration. And force at each point will be onky different by the same $-ag$ and net force unchanged.
Pendulum
Frequency changes. Normally the frequency for a pendulum is given by
$$f_p= \frac{\sqrt{g}}{2\pi \sqrt{L}}$$
Where $g$ is the salient factor because the inertial mass (resistance to acceleration) and the gravitational mass (provider of gravitational force) are the same and the $m$‘s cancel.
It would be tempting to think the new frequency is just
$$f_p= \frac{\sqrt{g-a}}{2\pi \sqrt{L}}??$$
But we can’t assume inertial and gravitational mass changes by the same proportion $\frac{a}{g}$. They don’t. Inertial mass unchanged. We could derive the frequency but hopefully you’re convinced that, in the case of the pendulum, you’re correct: Yes the oscillation frequency changes.
Edit: Calculating the Change
The frequency $\omega=2 \pi f$ for an oscillating system comes from:
$$\omega = \sqrt{\frac{k}{m}}$$
For low angle, $sin(\theta) \approx \theta$ is very close, and the restoring force perpendicular to the string is $F=-mgL~sin(\theta) \approx -mgL \theta$, giving constant $k=mgL$, using $\theta$ as our $x$ displacement ($F=-kx$). But the $k’$ for the accelerating frame case will not be the same.
New restoring force with accelerating elevator means:
$$k’=m(g-a)L=\frac{g-a}{g} ~ k \implies \omega’=\sqrt{\frac{g-a}{g}} \omega \implies f’=\sqrt{\frac{g-a}{g}} f$$
So the frequency will decrease if the elevator is accelerating up and increase if accelerating down. Fir example: if it’s accelerating up at 10% of $g$, the frequency will be $\sqrt{0.9}$ as much, down $5$%.
