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Question: For a simple pendulum, relation between time period \(T\), length of pendulum \(l\) and acceleration...

For a simple pendulum, relation between time period TT, length of pendulum ll and acceleration due to gravity ggis given by:
A. T=2πlgT = 2\pi \dfrac{l}{{\sqrt g }}
B. T=2πlgT = 2\pi \sqrt {\dfrac{l}{g}}
C. T=2πlgT = 2\pi \dfrac{l}{g}

Explanation

Solution

The time period of the simple pendulum is defined as the time required to complete one complete cycle of oscillation.
Restoring force fe{f_e} for the simple pendulum is:
fe=mgsinθ{f_e} = - mg\sin \theta
Here, mmmass of the suspended particle, ggacceleration due to gravity and θ\theta is angular displacement from the me
a position.
For simple harmonic oscillation acceleration aa is given as,
a=ω2xa = - {\omega ^2}x
Here, ω\omega is frequency of oscillation and x is the distance of mass from the attachment of the string.

Complete step by step answer:
It is given that the time period of oscillation is TT, the length of the pendulum ll and the acceleration due to gravity gg.
You have to find the relation between these quantities.
First we will calculate the time period of the simple pendulum using the formula given below,
T=2πωT = \dfrac{{2\pi }}{\omega } ….. (1)
You have to determine this ω\omega .

Figure: simple pendulum.
Consider a simple pendulum (figure above) a small angular displacement θ\theta . The cosθ\cos \theta component of mgmgis balanced by the tension of the string. And the sinθ\sin \theta component of the mgmg work as restoring force so,
fe=mgsinθ{f_e} = - mg\sin \theta .
Here, θ\theta is very less so sinθθ\sin \theta \approx \theta
θ\theta for this pendulum is given as θ=xl\theta = \dfrac{x}{l}, xx is distance of mass from the string is attached.
Using the above equation,
fe=mg(xl){f_e} = - mg(\dfrac{x}{l})
The acceleration is given as a=fema = \dfrac{{{f_e}}}{m},
a=g(xl)a = - g\left( {\dfrac{x}{l}} \right) …… (2)
For simple harmonic oscillation acceleration aa is given as,
a=ω2xa = - {\omega ^2}x …... (3)
Compare equation 1 and 2,
g(xl)=ω2x- g\left( {\dfrac{x}{l}} \right) = - {\omega ^2}x
Now, solve for ω\omega ,
ω=gl\omega = \sqrt {\dfrac{g}{l}} …… (4)
Put value of ω\omega from equation 4 to equation 1,
T=2πlgT = 2\pi \sqrt {\dfrac{l}{g}} .

So, the correct answer is “Option B”.

Note:
Always choose a coordinate system for such a problem and all measurements should be made from that coordinate system. First find the force acting on mass then make components of that force and equate the force to get the equation of motion.