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Discuss the simple pendulum in detail.

Answer» <html><body><p></p>Solution :Construction : A pendulum is mechanical system which exhibits periodic motion. It has a bob with mass m suspended by massless and inextensible string. The other end is fixed on a stand as shownin Figure. (a). At equilibrium, the pendulum doesnot oscillate and is suspended vertically downward. Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released. The bob of the pendulm executes to and fro motion. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C10_E01_035_S01.png" width="80%"/> <br/> Caclulation of time period. Let l be the length of thependulum which is taken as the distance between the point of suspension and the <a href="https://interviewquestions.tuteehub.com/tag/centre-912170" style="font-weight:bold;" target="_blank" title="Click to know more about CENTRE">CENTRE</a> of gravity of the job. <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> forces act on the bob of the pendulum at any displaced position, as shown in the figure (d), <br/> (i) The gravitational force acting on the body. `(<a href="https://interviewquestions.tuteehub.com/tag/vecf-3257300" style="font-weight:bold;" target="_blank" title="Click to know more about VECF">VECF</a>= mvecg)` that acts vertically donwards. <br/> (ii) The tensionin the string `vecT` that acts along the string to the point of suspension. <br/> Resolving the gravitational force into its components : <br/> (i) Normal <a href="https://interviewquestions.tuteehub.com/tag/component-926634" style="font-weight:bold;" target="_blank" title="Click to know more about COMPONENT">COMPONENT</a> : It is along the string but in opposition to the direction of tension `F_(as) = mg cos theta`. <br/> (ii) Tangential component : It is perpendicular to the string i.e., along tangential direction of are of swing `F_(ps)= mg sin theta`. <br/>Hence , the normal compoent of the forece, is along the string, <br/> `T-W_(as)=m(v^(2))/(l)` <br/> Here v is speed of bob <br/> `T- mg cos theta=m (v^(2))/(l) ""...(1)` <br/> From the figure, it is observed that the tengential component `W_(ps)` of the gravitationalforce always points towards the equilibrium position. This direction always points the bob from the mean position. Hence, in this case, the tangential force is the restoring force. Applying Newton's second law along tangential directiojn, we have <br/> `m(d^(2)s)/(dt^(2))+F_(ps)=0 rArr m=(d^(2)s)/(dt^(2))=-F_(ps)` <br/> `m(d^(2)s)/(dt^(2))=-mg sin theta ""...(2)` <br/> Where s is the position of bob that is measured along the <a href="https://interviewquestions.tuteehub.com/tag/arc-381236" style="font-weight:bold;" target="_blank" title="Click to know more about ARC">ARC</a>. Expressing arc length in terms angular displacement i.e. <br/> `s= l theta ""......(3)` <br/> Then its acceleration, <br/> `(d^(2)s)/(dt^(2))=l (d^(2) theta)/(dt^(2)) ""....(4)` <br/> Substituting equation (4) in equations (2), we get, <br/> `l(d^(2)theta)/(dt^(2))=-g sin theta""...(5)` <br/>Because of the presence of `sin theta` in the above differential equation, it is a non-linear differential equation. it is assumed that '' the small oscillation approximation,'' `sin theta ~~ theta`, the above differential equation becomes linear differential equation.<br/> `(d^(2)theta)/(dt^(2))=-(g)/(l)theta` <br/> It is known as oscillatory differential equation. Hence, the angular frequency of this oscillator (natural frequency of this system ) is<br/> `omega^(2)=(g)/(l) ""....(6)` <br/> `rArr omega= sqrt((g)/(l)" in rad" s^(-1) ""...(7)` <br/> The frequecny of oscillations is<br/> `f=(1)/(2pi) sqrt((g)/(l)) "in" Hz ""....(8)` <br/>and time period of oscillations is<br/> `T=2pi sqrt((l)/(g))` in second `"".....(9)` <br/></body></html>


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