# What is the meaning of “kay effective” $k_{\rm eff}$ in SHM?

I am really confused studying for my Physics lectures on oscillations, namely Simple Harmonic Motion. You see, my Professor introduced the topic: when he solved some examples, I noticed that when working with springs, he either:

• Uses the actual $k$ of the spring that is given in the wording of the exercise.

• Even though the exercise might give the value of $k$ for the spring, he sometimes works with it to find what he calls $k_{\rm eff}$.

I really do not understand how to tell each case apart and I am really desperate.

One example problem he solved in class is the following. Problem 4 of this problem set: http://web.mit.edu/8.01t/www/materials/ProblemSets/Raw/f12/ps11sol.pdf

I understand how it solves the problem, but how do I know I need to look for another expression of $\omega$ when, in general, we have that $\omega =\sqrt \frac km$, and k is already given in the problem?

I hope my question is understandable. It might sound like I have no idea, but I feel that I really do not in this new topic and I do not know where to start to understand this.

All help is greatly appreciated.

$k$ is the spring constant experienced by the body due to actual string. $k'$ effective is the constant in the expression $m\frac{d^2x}{dt^2} = - k'x$ after you've done some rearrangements to your differential equations. For example, supposes you have a mass on two springs of spring constants $k_1$ and $k_2$. Then, your differential equation is: $$m\frac{d^2x}{dt^2} = -k_1x - k_2x= -(k_2+k_1)x$$ where x is displacement from equilibrium. Then $k' = k_2 + k_1$. So although you might have two springs acting on a system, you get behaviour effectively equivalent to a single spring of spring constant $k'$. Now you can apply this idea to more complicated cases where you have different forces acting on the object. Important point is that if you can rearrange the equations into the $$m\frac{d^2x}{dt^2} = something * x$$ form, your something is the effective spring constant and you can treat the problem as a mass on a spring.
• Thanks so much. This is the first time I have seen an explanation that is so straightforward. I am wondering though, in class, we saw that when you have two strings in parellel, $K_{eff} = K_1 + K_2$, and when they are in series, $\frac 1{K_{eff}} =\frac 1{K_1} + \frac 1{K_2}$ Is this okay? – Bee May 14 '17 at 18:17
We have $$\omega=\sqrt{\frac{k}{m}}$$ only when the only force acting on the object is $$F=-kx$$ In your case, there is also friction and so it does not hold.