Name: Potential Energy of Simple Harmonic Motion
Uploaded: 2019-07-19
Duration: 7 min 29 s
Description: In simple harmonic motion, there is a continuous interchange of kinetic energy and potential energy. Let's learn to calculate the potential energy in this vi...

Question

. Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in figure.

Accepted Answer

For the calculation purpose, in this situation we will neglect gravity because it is constant throughout and will not affect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount $x_{0}$

Tension in the spring $= k x_{0}$

For the equilibrium of the mass $M$ , $mg = 2 k' x_{0} — (i)$

Let the mass be pulled through a distance $y$ , the string is extensible, hence spring alone will contribute to the extension $y + y = 2 y$ , to lower the mass down by $y$ from the initial equilibrium mean $x_{0}$ . So, next extension will be $2 y + x_{0}$ .

$mg - 2 k' (x_{0} + 2 y) = F mg - 2 k' x_{0} - 4 k' y = ma$

$mg - 4 k' y - mg = F$ $[$ Using equation $(i)]$

$\Rightarrow F = - 4 k' y$

Now since the net force $F$ is proportional to $- y$ , the system performs a simple harmonic motion.

Comparing the above net force with $F = - kx$ , we get

$k = 4 k'$

For simple harmonic motion, time period is given by

$T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{m}{4 k'}} = \frac{2 π}{2} \sqrt{\frac{m}{k'}} T = π \sqrt{\frac{m}{k'}}$

So, the time period of mass $M$ is $π \sqrt{\frac{m}{k'}}$ .

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