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Write down the time period of simple pendulum. |
Answer» Solution :A pendulum is a mechanical system which exhibits periodic motion. Ithas a bob with mass m suspended by a long string (assumed to be massless and inextensiblestring) and the other end is fixed on a stand as shown in Figure (a). At equilibrium, the pendulum does not oscillate and hangs verticallydownward. 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 pendulum executesto and fro motion. Let 1 be the length of the pendulumwhich is TAKEN as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position, as shown in the Figure (d),  Normal component: The componentalong the string but in opposition to the direction of tension, `F_("as")=mg cos theta`. Tangential component: The componentperpendicularto the string i.e., alongtangential direction of arc of swing, `F_(ps)=mg SIN theta`. Therefore, the normal component of the force is , along the string, `T-W_("as")=m(v^(2))/(l)` Here v is speed of bob `T-mg cos theta=m(v^(2))/(l)"" ...(1)` From the Figure we can observe that the tangential component`W_(ps)` of the gravitationalforce always pointstowards the equilibrium positioni.e., the direction in which it always points OPPOSITE to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton's second law along tangentialdirection, we have `m(d^(2)s)/(dt^(2))+F_(Ps)=0 rArr m (d^(2)s)/(dt^(2))= -F_(Ps)` `m(d^(2)s)/(dt^(2))= -mg sin theta "" ...(2)` where, s is the position of bob whichis measured along the arc. Expressing are length in terms of angular displacement i.e., `s=l theta ""...(3)` then its acceleration, `(d^(s))/(dt^(2))=l(d^(2)theta)/(dt^(2))""...(4)` Substituting equation (4) in equation (2), we get `l(d^(2)theta)/(dt^(2))= -g sin theta` `(d^(2)theta)/(dt^(2))=-(g)/(l)sin theta ""...(5)` Because of the presenceof `sin theta` in the above differentialequation, it is a non-linear DIFFERENTIAL equation (Here, homogeneous second order). Assume "the small oscillation approximation", `sin theta ~~theta`, the above differentialequation BECOMES linear differential equation. `(d^(2)theta)/(dt^(2))= -(g)/(l) theta""...(6)` This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is `omega^(2)=(g)/(l)rArr omega=sqrt((g)/(l))" in rad"s^(-1)` The frequency of oscillations is `f=(1)/(2pi) sqrt((g)/(l)) ` in Hz and time period of oscillations is `T=2pi sqrt((l)/(g))` in second. |
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