Infinite length Simple Pendulum

What is the time period of a simple pendulum of infinite length.

The answer is 86 minutes.

Can anyone explain this?

#Physics

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Comments

Time period of a long pendulum is given by $\frac { 2\pi }{ \sqrt { \left( \frac { 1 }{ R } +\frac { 1 }{ l } \right) g } }$
for infinite pendulum 1/l will be neglected and the formula will be $2\pi \sqrt { \frac { R }{ g } }$
which on solving results in 86 mins

Rohit Gupta - 6 years ago

bhai,how did you get this formula,proof??

Piyush Kumar Behera - 4 years, 4 months ago

What is R? Also, what if the pendulum is in a gravity free region? Won't the friction of pendulum parts slow it down? Will the time period be different for the types of material we use to make bob and the pendulum?

Please answer!

Kaushik Chandra - 3 years, 10 months ago

R is the radius of the Earth.

In gravity free region the pendulum won't oscillate at all.

Friction will slow down all the pendulums whether infinite or not does not matter.

The time period does not depend on the bob if it is small.

Rohit Gupta - 3 years, 10 months ago

Take any instant t let position at t=P join the mean position of the pendulums bob to the centre of the earth join P to the centre of the Earth Let angle between both lines be Q at P: mg acts vertically downwards mgsinQ provides restoring force F= -mgsinQ a=F/m= -gsinQ.........(1) for Q<10 degrees sinQ=Q(in radians) = tanQ let distance between the mean position and P be x tan Q=x/R where R is radius of earth a= -gx/R but a= -w^2 X x therefore w=underroot of g/R time period= 2pi/w therefore T= 2pi underroot (g/R) that is 86 Minutes

Arya Vardhan Singh - 3 years, 4 months ago

it is because t=2pisqrtradius of earth/g

Varun Khare - 7 years, 4 months ago

You should consider a pendulum with the length L which is much longer than the Earth radius R. If you then use for the description of the pendulum motion the conservation law in the gravitational field (not in the uniform field), you will get the following result - angular frequency to the second power will be equal to g/R.

Сергей Кротов - 7 years, 4 months ago

You are right, it is just the change of g in the points close to the surface that makes the system oscilate harmonically. You should consider the change of the gravitational energy when the displacement is small and equate it to the change of kinetic energy. Finally in this particular case when the body is close to the surface of the Earth (in the strong limit L>>R) the period wouldn't depend on L.

Сергей Кротов - 7 years, 4 months ago

Ans is 84.6 minute

Abi Singla - 4 years, 4 months ago

t = 2 * pi * sqrt( l / g )

so i think if the its length is infinite so the time period also will be infinite :D

i don't know whether i am right or wrong :D

Sherif Elmaghraby - 7 years, 4 months ago

Can someone please post a detailed proof??

Amlan Mishra - 5 years, 3 months ago

i have posted a detailed proof

Arya Vardhan Singh - 3 years, 3 months ago

http://www.quantumstudy.com/physics/simple-harmonic-motion-12/ Check out here

Narendra Bumb - 3 years, 3 months ago

THE LENGTH OF THE PENDULUM IS MUCH LONGER THAN THE RADIUS OF THE EARTH SO IN THE FORMULA OF THE TIME PERIOD WE CONSIDER THE CHANGE IN THE VALUE OF g.

Pranjal Rajawat - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$