Analysis of a spring-block motion when the point of suspension is accelerating vertically upward I was taught to apply a "pseudo force", [which is equal to mass × acceleration (of point of suspension)] in the direction opposite to the point's acceleration and then proceed the way I would have had there been no acceleration, only replacing g by g + a (since it is accelerating upwards). I don't understand the intuition behind this, so how can I obtain the equation of motion for the block without considering a pseudo force?
 A: Viewed from a non-inertial (accelerating) reference frame, the equation of motion for a system must account for the effects of the observer's acceleration.  The acceleration of the non-inertial frame can be translational and/or rotational.  See a physics mechanics text, such as Mechanics by Symon.
Your question deals with translational acceleration.  Consider a block of mass $m$ sitting on the ground; it is at rest with respect to an observer standing on the ground.  That is, the block is at rest in the inertial frame of the observer.  (I am neglecting the effects of the earth's rotation here.  The earth is actually a non-inertial frame due to its rotation.)  There are two forces on the block: $m \vec g$ downward and $\vec N$, the constraint from the earth, upwards.  In this frame, the equation of motion of the block is $m \vec g + \vec N = m \vec a = 0$, where $\vec a$ is the acceleration of the block in the inertial frame, zero for the stationary block.
Now, consider the motion of the block from the viewpoint of an observer in an elevator that is accelerating upwards with acceleration $\vec a'$.  Relative to this observer the block is accelerating downward with velocity $- \vec a'$, but there is no force in the inertial frame of the block that causes the apparent acceleration, it is solely due to the acceleration of the observer.  But since the observer sees the block accelerating, the observer attributes this to some force $-m \vec a'$, and in this non-inertial frame the equation of motion is $m \vec g + \vec N - m \vec a' = m \vec a^{*}$ where $\vec a^{*}$ is the acceleration in the non-inertial frame.  Since $m \vec g + \vec N = 0$, $- m \vec a' = m \vec a^{*}$; to this observer the block is accelerating at $- \vec a'$.  In the equation of motion in the non-inertial frame, there is a force that appears due to the acceleration $\vec a'$ of the reference frame, this force is $-m \vec a'$.  Since this is not a force in the inertial frame it is called a "fictitious" force because it only appears in the non-inertial frame.
(If the non-inertial frame is also rotating other fictitious forces arise, such as the centrifugal and Coriolis forces.)
Consider your specific situation for the the motion of a block mass $m$ held by a spring.  First, assume the block is in a box suspended by a spring hung from the top of the box with the box at rest in the inertial frame, and you wish to evaluate the motion from a frame with translational acceleration.  In the inertial frame the equation of motion is $m \vec g + \vec F_{spring} = m\vec a$ where $\vec a$ is the acceleration in the inertial frame. Viewed from a non-inertial reference frame that is in translational acceleration $\vec a'$ with respect to the inertial frame, in addition to the forces in the inertial frame $m \vec g + \vec F_{spring}$, there is also the fictitious force $-m \vec a'$ in the non-inertial frame so the equation of motion in the non-inertial frame is $m \vec g + \vec F_{spring} - m \vec a' = m \vec a^{*}$ where $\vec a^{*}$ is the acceleration as viewed from the non-inertial frame.  If $ \vec a'$ is up then $- \vec a'$ is downward, in the same direction as $\vec g$, so the effect is as if $g$ is increased to $g + a'$ in the non-inertial frame.
Now, assume the box is accelerating upwards in the inertial frame at $\vec a'$; there is an additional force on the mass in the inertial frame $\vec F_{up} = m\vec a'$ and the equation of motion for $m$ in the inertial frame is $m \vec g + \vec F_{spring} + \vec F_{up} = m\vec a$. $\vec F_{spring}$ is the spring force on $m$ for the box at rest in the inertial frame.  $F_{up}$ must be provided by the spring, so with the box accelerating in the inertial frame, the force of the spring on $m$ in the inertial frame is $\vec F_{spring} \enspace ' = \vec F_{spring} + \vec F_{up}$, and the equation of motion in the inertial frame is $m \vec g + \vec F_{spring} \enspace ' = m\vec a$ .  Viewed from a non-inertial frame accelerating at $\vec a^{''}$, the equation of motion in the non-inertial frame is $m \vec g + \vec F_{spring} \enspace' - m \vec a^{''} = m\vec a^{*}$ where
$\vec a^{*}$ is the acceleration in the non-inertial frame. If $ \vec a^{''}$ is up then $- \vec a^{''}$ is downward, in the same direction as $\vec g$, so the effect is as if $g$ is increased to $g + a^{''}$ in the non-inertial frame.  If
$\vec a^{''} = \vec a'$, the box is at rest in the non-inertial frame, and the equation of motion for $m$ in the non-inertial frame is $m \vec g + \vec F_{spring} = m\vec a^{*}$.  The spring forces on $m$ can be specified as follows. Let $\hat i$ be a unit vector in the upward direction, $k$ be the spring constant, and $x$ be the position of the weight on the spring measured downward from the top of the box, where the spring is secured.  Let $l$ be the value of $x$ when the string is relaxed (not stretched or compressed). $\vec F_{spring} = k(x-l) \hat i$. $\vec F_{up} = k(x' - l) \hat i = m  a' \hat i$; so $x' - l = ma'/k$. $\vec F_{spring} \enspace ' = k(x - l + (ma')/k) \hat i$. The equation of motion in the inertial frame is $-mg \hat i + k(x - l + (ma')/k) \hat i = ma \hat i = md^2x/dt^{2} \hat i$.  Let $X = x - l - mg/k + ma'/k$; then, the equation of motion becomes $md^2X/dt^2 + kX = 0$, which has solution $X(t) = c_1Cos(\omega _0t) + c_2Sin(\omega _0t)$, with $\omega _0 = \sqrt{k/m}$ where the constants $c_1$ and $c_2$ depend on the initial conditions for $X(0)$ and $\dot X(0)$.
The "fictitious" forces are "real" in the sense that they are "felt" in the non-inertial frame.  When your car accelerates forward, you feel the fictitious force pushing you back against the car seat.
A: Position to arbitrary point from reference frame $S$ can be composed of position from that frame to the origin of frame $S'$ and position from $S'$ to the arbitrary point.
More pictorially:

Or in vector notation:
$\vec r = \vec R + \vec {r'}$
Taking derivative w.r.t. time $(\frac{d^2}{dt^2})$ twice we end up with:
$\vec a = \vec A + \vec {a'}$
I.e. acceleration of arbitrary point P, viewed from frame $S$ is equal to the composite of acceleration of frame $S'$ (viewed from frame $S$) + acceleration of point P viewed from frame $S'$
In general $\vec F = m\vec a$
Letting $S$ be inertial frame and $S'$ be non-inertial one and wanting to express motion of the object from the latter, we must have:
$$\vec {F'} = m\vec {a'} = m(\vec a - \vec A) = \vec F + \vec {F_{ficticious}}$$
What does it have to do with your example? Well it is quite a powerful tool (transformation), that lets you 'move' between reference frames.
For example in the problem you provided, viewed from the inertial $S$ frame, acceleration of an object of mass m is given by:
$$\vec a = \frac{\vec F_{net}}{m}$$
, where $\vec F{ext}$ are the forces acting on the object (all the good stuff, gravitational force, external forces etc. but none of that 'fictitious nonsense')
If you will, consider the same situation viewed from non-inertial reference frame $S'$:
Acceleration measured w.r.t. this frame is:
$$\vec {a'}=\vec a - \vec A = \frac{\vec F_{net}}{m} + \frac{\vec F_{ficticious}}{m}$$
(all the good stuff, gravitational force, external forces etc. with that 'fictitious nonsense')
Some questions you may want to consider:

*

*Can you explain why pseudo-force is acting opposite to the acceleration of the non inertial $S'$ frame?
(hint: see second provided equation for acceleration in different frames)

*Using the first equation, can you prove, that Newton's Second Law is the same in all inertial reference frames?

