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प्रश्न
Discuss the simple pendulum in detail.
उत्तर
Simple pendulum: A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be a massless and inextensible string) and the other end is fixed on a stand. At equilibrium, the pendulum does not oscillate and hangs vertically downward. Such a position is known as the 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 executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the center of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position.
(i) The gravitational force acting on the body `(vec"F" = "m"vec"g")` which acts vertically downwards.
(ii) The tension in the string `vec"T"` which acts along the string to the point of suspension.
Resolving the gravitational force into its components:
(a) Normal component: The component along the string but in opposition to the direction of tension, Fas = mg cos θ
(b) Tangential component: The component perpendicular to the string i.e., along the tangential direction of an arc of the swing, Fps = mg sin θ
Therefore, The normal component of the force is, along the string,
T − Was = `"m""v"^2/"l"`
Here v is speed of bob
T − mg cos θ = `"m""v"^2/"l"`
From the figure, we can observe that the tangential component Wps of the gravitational force always points towards the equilibrium position i.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 tangential direction, we have
`"m"("d"^2"s")/("dt"^2) + "F"_"ps" = 0 ⇒ "m"("d"^2"s")/("dt"^2) = -"F"_"ps"`
`"m"("d"^2"s")/("dt"^2)` = − mg sin θ ..........(1)
where s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
s = lθ ...........(2)
then its acceleration, `("d"^2"s")/"dt"^2 = "l"("d"^2θ)/"dt"^2` ............(3)
Substituting equation (3) in equation (1), we get
`"l"("d"^2θ)/"dt"^2` = − g sin θ
`("d"^2θ)/"dt"^2 = - "g"/"l" sin θ` .........(4)
Because of the presence of sin θ in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ≈ θ, the above differential equation becomes a linear differential equation
`("d"^2θ)/"dt"^2 = - "g"/"l" θ` ...........(5)
This is the well-known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is
`ω^2 = "g"/"l"` ..........(6)
∴ `ω = sqrt("g"/"l")` in rad s−1 .......(7)
The frequency of oscillations is f = `1/(2π) sqrt("g"/"l")` in Hz .........(8)
and time period of oscillations is T = `2π sqrt("l"/"g")` in second .........(9)
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