Consider a block of mass $m$ hunged by a spring of spring constant $k$. Let it be in equilibrium position where $mg=kx$.

Now it is given a sudden impulse in upward direction so that it starts moving in upward direction, does the maximum height it would reach will be calculated as $mv^2/2=mgh$ where $h$ is the max height reached? Or will the height reached will be different as the spring will also exert force and thus involve itself in energy conservation equation? The above equation I think is only valid if there is no spring force throughout whole journey, but this does not sit well with my intuition. So what is actually happening?


2 Answers 2


This problem is best addressed via energy. With $z$ as the height above the equilibrium position of the unloaded spring, the loaded system has potential:

$$ U(z) = \frac 1 2 kz^2 + mgz $$

The equilibrium position is:

$$ z_0 = -g\frac m k $$

so that:

$$ U(z_0)= \frac 1 2 kz_0^2 + mgz_0= -\frac 1 2 \frac{m^2g^2}{k}\equiv U_0$$

Now it's given an impulse. The details don't matter, all we need to know is the kinetic energy:

$$T_0= \frac 1 2 mv^2 $$

With that, you have the initial total energy:

$$ E_0 =U_0+T_0 = \frac 1 2 mv^2 - \frac 1 2 \frac{m^2g^2}{k} $$

$$E_0 =\frac m 2 \Big( v^2 - \frac{g^2}{\omega_0^2} \Big)$$

where $\omega_0=\sqrt{k/m}$.

As the mass moves:

$$ U(z)+\frac 1 2 m\dot z^2=E_0$$

which can solved for $z$ with $\dot z=0$.

  • $\begingroup$ Thanks and i would like to ask that if the mass is going upwards then it will also cross the equilibrium point and then from there compression of the spring will start, so where is the term that is addressing this factor. And more more thing, after plugging some values i got the U(z) in negative, so what does it mean. $\endgroup$
    – Cyberax
    Oct 28, 2021 at 4:03
  • $\begingroup$ That's exactly why I chose the coordinate zero where I did. All linear dependence is gravity, while all the spring dependence is quadratic. Any other choice won't separate them exactly. $\endgroup$
    – JEB
    Oct 28, 2021 at 18:42

At equilibrium, the spring force balances gravity, such that the behavior of the system is equivalent to that of mass $m$ and spring $k$ without gravity, just shifted by $x$. The maximum height (relative to equilibrium position) can be found by exchanging the initial kinetic energy with that absorbed by the spring. $g$ need not appear. In other words: $$\frac12mv^2 = \frac12kh^2$$

Or, keep gravitation potential energy in as well as spring energy; just take care that zero spring energy is placed at $x$ above equilibrium. Then, substitutions using the equilibrium statement will allow terms involving gravity to cancel each other out.


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