# Oscillation of the plank

1 vote
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In the figure shown a plank of mass M is placed on a smooth horizontal surface . Two light identical springs of stiffness k each are rigidly connected to struts at the ends of the plank as shown in the figure

When the springs are in their unextended positions the distance between their free ends is 3l. A block of mass m is placed on the plank and pressed against one of the spring so that it is compressed by a length I. To keep the block at rest it is connected to the strut by means of a light string as shown in the figure. Initially,the system is at rest and the string is burnt through. Then (ignore any friction.) we have to find the period of oscillation of the plank .

I tried as

Now how to proceed , the answer given as

edited Jan 24, 2017

The CM of the plank and block must remain stationary, because there are no net external forces acting. So the period of $M$ is the same as that of $m$.
I don't understand where the answer you have been given is coming from.

Possibly the block $m$ is not attached to the springs, so the springs only exert a force on $m$ when they are compressed. If so, your question needs to be altered. I think you have missed out several words or phrases, some of which are crucial to understanding the problem.

Please edit your question to show the **exact wording** of the problem you have been given.

Assuming that the springs do not extend, the next step is to find the velocity of the block $m$ relative to the plank as it travels between the springs. If the time taken to travel this distance is $t_2$ then the period of the motion is $2(2t_1+t_2)$.
Sorry for the inconvenience .  This is the exact question http://dc.allenbpms.in/testpaper/solution/d8676-28-636206202521737165-028s.gif
Whoever photocopied the question missed some text on the right. Perhaps missing text is also preventing you from understanding the solution. Do you understand the problem now? Is there anything in the given solution which you still do not understand?
Yes the steps done after we got $t_1$
Can you be more specific?
I just don't get how they have written this  http://i.imgur.com/R0WbjOS.jpg
But your answer given has good explanation , that has cleared my doubt

1 vote

The oscillation of the plank and block can be considered in 2 parts.

In the SHM part of the motion the block is in contact with one of the 2 springs. The springs are never extended, only compressed, but as one spring is compressed the forces between the block and plank are the same as if the other spring were extended. So ignoring what happens in the gap, the system is equivalent to a two mass one spring system. The angular frequency of this motion is given by $\omega^2=\frac{k}{\mu}$ where $\mu=\frac{mM}{m+M}$ is the reduced mass. The period of this oscillation is $T=\frac{2\pi}{\omega}$.

In the other part of the motion the block and plank are moving towards each other with constant velocity, starting from a distance $3\textit{l}$ apart, because that is the separation of the relaxed springs. (I assume the block has negligible width, otherwise this separation must be reduced by the width of the block.) There is no PE stored in the spring at this stage, so the total KE now equals the initial PE :
$\frac12k\textit{l}^2=\frac12mv^2+\frac12MV^2$.

Momentum is conserved so $MV=mv$ hence $MV^2=(\frac{m}{M})mv^2$. The relative velocity of approach between block and the far spring is
$v+V=v(1+\frac{m}{M})=mv/\mu$.
Therefore
$k\textit{l}^2=(1+\frac{m}{M})mv^2=m^2v^2/\mu$
$(mv)^2=\mu k \textit{l}^2=\frac{k}{\mu}(\mu\textit{l})^2$
$mv=\omega \mu\textit{l}$.

The time taken to cover the round-trip distance of $2\times 3\textit{l}$ between the springs is
$t=\frac{6\textit{l}}{v+V}=\frac{6\textit{l} \mu}{mv}=\frac{6\textit{l} \mu}{\omega \mu \textit{l}}=\frac{6}{\omega}$.

Therefore the total period of the oscillation is $T+t=(2\pi+6)\frac{1}{\omega}$.

answered Jan 24, 2017 by (26,096 points)
edited Jan 24, 2017