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Consider a simple pendulum, having a bob attached to a string , that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple. Pendulum depends on its length ( l) , mass of the bob ( m ) , and acceleration due to gravity ( g ) . Derive theexpression for is time period using method of dimension .

Answer» <html><body><p></p>Solution :(i) Let `theta` be the angle made by the string with vertical. When the bob is at the mean position `theta=0.`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V02_C10_E15_008_S01.png" width="80%"/><br/> (ii) There are two forces acting on the bob. The tension T along the vertical force due to gravity (=mg).<br/> (iii) The force can be resolved into `mgcostheta` along a circle of <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> L and centre at this support point. <br/> (iv) Its radial <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> `(omega^(2)L)` and also tangential acceleration. So net radial force `= T - mg costheta` , While the tangential acceleration provided by `mg sintheta`the radial force gives zero torque. <br/> (v) So torque providedby the tangential component. <br/> `tau="Lmg"sintheta` <br/> `tau=Ialpha` (by Newton's law of rotational motion) <br/> `Ialpha=mgsinthetaL` <br/> `alphaimplies("mgL")/(I)sintheta` <br/> where I is the <a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> of inertia , `alpha` is angular acceleration, <br/> (<a href="https://interviewquestions.tuteehub.com/tag/vi-723586" style="font-weight:bold;" target="_blank" title="Click to know more about VI">VI</a>) The `theta` is to small, <br/> `sintheta=theta` <br/> `alpha=(mgL)/(I)theta,omega=sqrt((I)/(mgL))`<br/> [for simple <a href="https://interviewquestions.tuteehub.com/tag/harmonic-1015999" style="font-weight:bold;" target="_blank" title="Click to know more about HARMONIC">HARMONIC</a> `a(t) =-omega^(2)txxtheta" is small ]"and T=2pisqrt((I)/(mgL))""[:'omega=(2pi)/(T)]`</body></html>


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