question_answer

From the variation of potential energy in the direction of small oscillation of a simple pendulum, find the effective spring constant for the simple pendulum, where m is mass of the bob, \[l\] is length of the simple pendulum.

A) \[\frac{mg}{l}\]            

B) \[\frac{mg}{2l}\]

C) \[\frac{2mg}{l}\]          

D) \[\frac{mg}{\sqrt{2}l}\]

Correct Answer: A

Solution :

[a] When a particle is executing SHM, at mean position \[F=-\frac{dU}{dr}=0\]Also, its potential energy is minimum at equilibrium position. So.\[\frac{{{d}^{2}}U}{d{{r}^{2}}}>0\]. \[\frac{dU}{dr}=-F\Rightarrow \frac{{{d}^{2}}U}{d{{r}^{2}}}=-\frac{dF}{dr}\] At equilibrium, \[dF=-{{\left| \frac{{{d}^{2}}U}{d{{r}^{2}}} \right|}_{r=0}}dr\] Comparing this standard SHM equation, i.e. \[{{F}_{net}}=-{{K}_{eff}}x,\]We have \[{{d}_{eff}}={{\left| \frac{{{d}^{2}}U}{d{{r}^{2}}} \right|}_{r=0}}\] In general equilibrium happens at \[r={{r}_{0}}\](say) \[\therefore {{k}_{eff}}={{\left| \frac{{{d}^{2}}U}{d{{r}^{2}}} \right|}_{r={{r}_{0}}}}\] Also, at equilibrium \[F=-\frac{dU}{dr}=0\] Now, coming to the situation given, when the bob id given a small horizontal displacement as shown. \[U=mgy\] \[y=l(1-cos\theta )=2l{{\sin }^{2}}\frac{\theta }{2}\] \[\therefore U=2mgl{{\sin }^{2}}\frac{\theta }{2}\] So \[U=\frac{1}{2}mgl\left( \frac{\theta }{2}.\frac{\theta }{2} \right)\]or \[U=\frac{1}{2}mgl{{\theta }^{2}}\] \[\theta \simeq \frac{x}{l}\Rightarrow U=\frac{mg}{2l}{{x}^{2}}\] \[\frac{dU}{dx}=\frac{mg}{2l}.2x=\frac{mg}{l}.x\] \[\Rightarrow \frac{{{d}^{2}}U}{d{{x}^{2}}}=\frac{mg}{l}\]\[\therefore {{k}_{eff}}={{\left| \frac{{{d}^{2}}U}{d{{x}^{2}}} \right|}_{x=0}}=\frac{mg}{l}\]