Problem 12 of section B of this PDF file reads

Two springs with spring constant k1 and k2 are attached to a body of mass (m) in two different configurations in 2 cases (A and B) as shown.


The mass rests on a surface with coefficient of friction between the mass and surface as $K$. The mass is moved to the right, parallel to the axis of the springs by a distance $d$ (from natural lengths of the springs) and is released. The time taken till the body comes to rest in case A is:

The given answer is equal to B but I get it to be less than B.

$$T= 2\pi \sqrt{\frac{m}{k}} $$

But I'm not sure how to incorporate effect of friction and mass. In case of A, there is a $$ frictional force F = Kmg $$

But how do I relate this to T ? I referred the pdf link above. I was trying to solve the question. The answer is $$ T_{a} = T_{b} $$

An alernate way that I just figured is given below. Please clarify -
$For A $ ,$$ -k_{1}d - k_{2}d - Kmg = m\omega_{a} ^2 x $$ $For B$, $$ -k_{1}d - k_{2}d - Kmg = m\omega _{b}^2 x $$ $$ \omega_{a} = \omega_{b} $$ $$ T=\frac{2Pi}{\omega} $$ $$ T_a=T_b $$ I am not a physics student. So I wanted to understand this conceptually and get the answer from the relevant equations. Also is my approach to the problem correct ?

  • 1
    $\begingroup$ Hi coolstuff - I edited your question to clean it up a bit. If it doesn't reflect your original intent, please edit it further so that it does. On a separate note, did you look in a textbook or on the internet to see if you could find anything about this? See our suggestions of prior research here. $\endgroup$
    – David Z
    Commented Dec 6, 2014 at 9:41
  • $\begingroup$ @DavidZ Yes , I did do prior research. That's how I came to some conclusions. But I wasn't sure about mass and friction here . $\endgroup$
    – coolstuff
    Commented Dec 6, 2014 at 9:49
  • 2
    $\begingroup$ In that case I'd suggest editing the post to explain some of what you tried. In particular, many resources will show you how to incorporate the effects of friction and mass in a spring system, so anyone reading this question is going to wonder why you didn't just look at those resources. A good question addresses those concerns. Tell us some of the resources you looked at, what you tried after looking at them, and why it didn't work out. You don't have to include everything, but it really helps to show some of the dead ends you hit on the way to posting your question. $\endgroup$
    – David Z
    Commented Dec 6, 2014 at 9:52
  • $\begingroup$ @DavidZ made edits. $\endgroup$
    – coolstuff
    Commented Dec 6, 2014 at 11:14
  • $\begingroup$ OK, the part you added about the frictional force helps a lot. The part about the alternate way you found to solve the question doesn't help, IMO, because it's not clear how it relates to the main question you're asking (namely, how to account for friction and mass when solving the problem). It starts to sound like you're asking why the two approaches give different answers, which is a different question. $\endgroup$
    – David Z
    Commented Dec 6, 2014 at 11:31

2 Answers 2


You actually do not need to solve the equation at all in order to determine that, whatever the answer, it must be the same for A and B.

The only difference between A and B is the direction of k1. However if you look at $F=-kx$ is always a restoring force whose magnitude is proportional to the magnitude of the displacement from equilibrium. So the mass always experiences a restoring force of the same magnitude regardless of whether k1 is being compressed or stretched.

So in both cases the forces are the same and therefore the time must be the same also.

  • $\begingroup$ What about the frictional force mentioned in the question ? Also in A, k2 is compressed,k1 is stretched. In B, k1 and K2 are both compressed. So wouldn't that matter ? $\endgroup$
    – coolstuff
    Commented Dec 6, 2014 at 12:26

Since both springs are connected effectively in parallel, the both cases will have a same equation of motion. The equation of motion: $ma = F$

$$ \tag{1} m \frac{d^2x}{dt^2} = -2kx - f_d. $$

The variable $x$ is the displacement for the oscillator displacement from its equilibrium position, $m$ the mass, $f_d$ is the dragging or frictional force. It is not easy to model a constant static frictional force, because it is oscillation motion, the displacement $x$ changes sign regularly. It is more convinient to model a dragging force propotional to velocity, $f_d = - b v$, the velocity will take care the sign of the force. Therefore, lets adopt this dragging force model:

$$ \tag{2} m \frac{d^2x}{dt^2} = -2kx -b v; $$ where $b$ is the damping coefficient. Write Eq. (2) in standard form of a second order differential equation: $$ \tag{3} m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + 2kx= 0; $$ Scale the equation by mass $m$, and define $\omega_0 = \sqrt{\frac{2k}{m}}$ , $\gamma = \frac{b}{2m}$. We then write the equation as:

$$ \tag{4} \frac{d^2x}{dt^2} + 2\gamma \frac{dx}{dt} + \omega_0^2 x = 0; $$

Assume that $\gamma \lt \omega_0$ (the underdamping regime). The solution for Eq. (4):

$$ x(t) = e^{-\gamma t} \left\{ A \cos (\omega t) + B \sin(\omega t) \right\}. $$ Where $\omega = \sqrt{\omega_0^2 - \gamma^2}$, A and B are two constnats to be determined by the initial condition: $x(0)$ and $v(0)$, the initial displacement and initial velocity.


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