Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Time period in different cases

1 vote

In this how they hwve written $T_1$
for simple pendulum I know $T=2\pi \sqrt\frac{l}{g}$

I also could not understand how they have written $F_2$

asked Mar 25, 2017 in Physics Problems by koolman (4,286 points)

1 Answer

2 votes
Best answer

1st case:

The complete formula for time period of simple pendulum is $ T = 2\pi \sqrt {\frac {1}{g (1/L + 1/R)}} $
Its derivation can be seen here

But when L << R then the formula reduces to $T=2\pi \sqrt {L/g} $

2nd case :
you may draw a circle as earth take center O . Make a tunnel along diameter. Let a body fall along tunnel and has displacement y. As the body approaches center the force decreases because here we know force acts on a particle at a distance y from its center is only due to the part of earth inside and outer part does not exert any net force.
Hence $ F= -GM' m/ y^2$ -----1

M= total mass of earth. M' = effective mass of earth.

$M' = \rho 4/3 \pi y^3$

so $M'/M = y^3/R^3$

substitute M' in equation 1 .

$F=-( GmM/R^3 )×y$

Compare it with$ F=-kx $ take out$ k$ and finally put it into:

$2\pi \sqrt {m/k } $
You will get $T=2\pi \sqrt {R/g}$.

3rd case
$ mw^2R = mg $ on solving you would get :
$T=2\pi \sqrt {R/g}$.

update : m is mass of pendulum.

answered Mar 25, 2017 by physicsapproval (2,320 points)
selected Mar 26, 2017 by koolman
From this "the part of earth inside and outer part "
What do you mean by outer and inner part . And 'part' of what ?
lets take example from an analogy of electrostatics, when we applied gauss law on a non conducting sphere to take out foeld at an "inside point" i.e. less than its radius we take only the field due to charge only inside same is happening here .
Case 1: Where does the "complete formula" come from? What does it mean?
What do you mean by effective mass of earth and what is m ?
By complete I meant that$ 2\pi \sqrt {l/g} $ was a result of that.  Its derivation can be seen on :
@koolman  the concept I used is same as when we calculate acceleration due to gravity within the earth.
A better explanation is here http://physics.stackexchange.com/questions/99117/why-gravity-decreases-as-we-go-down-into-the-earth
I have adapted some part of answer from the link I provided.

It's actually not entirely true that the strength of the Earth's gravitational field decreases as a function of depth. It is true for certain regions in the Earth, but it's untrue for others because of the non-trivial dependence of the Earth's density on depth.

To see what's going on, assume that the Earth is a sphere whose density is spherically symmetric.

Now consider a mass mm at some radius rr from the center of the Earth. Using Newton's Law of Gravitation, one can show that that given the spherical symmetry, the gravitational attraction on mm of all mass with radii greater than rr exert no net force on it. It follows that only the mass with radii less than or equal to rr contribute to the gravitational force on m
@Koolman , are you clear Or should I edit my answer ?
It is koolman's question, so it is for him to decide if your explanation is good enough. It looks ok to me.
Its good enough to understand .