What is the time period at the centre of earth of a bob 1 kg?
Answers
Answer:
For the case of the pendulum that was shown, this comes down to:
−FT=mLd2θdt2
For the force FT
, we know that it is a projection of the gravitational force Fg. We can determine the factor the gravitational force should be multiplied with easily. The vertical line together with the pendulum cord and the red line make for an isosceles triangle. Determining the angle of one of the equal angles, we get an angle of π−θ2
. Projecting the gravitational force correctly, we now get the updated equation:
−Fgsin(π−θ2)=mLd2θdt2
It now rests us to find a correct expression for the gravitational force. The general formula for the gravitational force is given by:
Fg=Gm1m2r2
where G
is the gravitational constant, m1 and m2 are the masses of the objects you calculate the force between, and r
is the distance between those two objects.
Since we are in the centre of the Earth however, we can’t just take the full mass of the Earth in the equation. We therefore must make use of the shell theorem (Shell theorem - Wikipedia). It states that only the mass within the distance of the pendulum and the centre is relevant to the problem. For this reason, we must now make another assumption. This assumption is that the mass of the Earth has a constant density ρ
. We now get for the masses in the gravitational equation mentioned before that:
m1=m
m2=Vρ=43πρr3
where r
is the distance from the pendulum to the centre of the Earth, denoted in red in the scheme that was presented earlier. When we take these masses into account, we get an updated gravitational equation for this particular case that reads:
Fg=43mGπρr
Everything is constant in this equation except for the distance r
. Let us therefore express this distance as a function of θ. This can be done by observing that this distance is actually equal to the length of a chord. For an angle θ, this length is equal to 2Lsin(θ/2)
.
We thus get a gravitational force that is equal to:
Fg=43mLGπρ⋅2sin(θ2)
If we put this back into the second equation of motion we originally had, we come to the equation we were looking for, which looks as follows:
−43mLGπρ⋅2sin(θ2)sin(π−θ2)=mLd2θdt2
This equation can be simplified by observing that the sine functions can be combined into one:
2sin(θ2)sin(π−θ2)=sin(θ)
Thus we get for our equation:
−43Gπρsin(θ)=d2θdt2
This should look very familiar to everyone who ever solved a harmonic oscillator. The solution method is the exact same. We assume only small deviations, which allows us to say sin(θ)≈θ
.
The pendulum will thus undergo simple harmonic motion, just like a regular simple pendulum. The solution to get to the period is straightforward now. It is given by (Thank you to Isaac Clark for pointing out that my previous expression didn’t have the right dimensionality since I wrote down the solution for the angular frequency.):
T=2π43Gπρ√=3πGρ−−−√
This period is, as you can see, independent of the mass of the pendulum, but also independent of the length of the pendulum cord.
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Explanation:
Answer:
For the case of the pendulum that was shown, this comes down to:
−FT=mLd2θdt2
For the force FT
, we know that it is a projection of the gravitational force Fg. We can determine the factor the gravitational force should be multiplied with easily. The vertical line together with the pendulum cord and the red line make for an isosceles triangle. Determining the angle of one of the equal angles, we get an angle of π−θ2
. Projecting the gravitational force correctly, we now get the updated equation:
−Fgsin(π−θ2)=mLd2θdt2
It now rests us to find a correct expression for the gravitational force. The general formula for the gravitational force is given by:
Fg=Gm1m2r2
where G
is the gravitational constant, m1 and m2 are the masses of the objects you calculate the force between, and r
is the distance between those two objects.
Since we are in the centre of the Earth however, we can’t just take the full mass of the Earth in the equation. We therefore must make use of the shell theorem (Shell theorem - Wikipedia). It states that only the mass within the distance of the pendulum and the centre is relevant to the problem. For this reason, we must now make another assumption. This assumption is that the mass of the Earth has a constant density ρ
. We now get for the masses in the gravitational equation mentioned before that:
m1=m
m2=Vρ=43πρr3
where r
is the distance from the pendulum to the centre of the Earth, denoted in red in the scheme that was presented earlier. When we take these masses into account, we get an updated gravitational equation for this particular case that reads:
Fg=43mGπρr
Everything is constant in this equation except for the distance r
. Let us therefore express this distance as a function of θ. This can be done by observing that this distance is actually equal to the length of a chord. For an angle θ, this length is equal to 2Lsin(θ/2)
.
We thus get a gravitational force that is equal to:
Fg=43mLGπρ⋅2sin(θ2)
If we put this back into the second equation of motion we originally had, we come to the equation we were looking for, which looks as follows:
−43mLGπρ⋅2sin(θ2)sin(π−θ2)=mLd2θdt2
This equation can be simplified by observing that the sine functions can be combined into one:
2sin(θ2)sin(π−θ2)=sin(θ)
Thus we get for our equation:
−43Gπρsin(θ)=d2θdt2
This should look very familiar to everyone who ever solved a harmonic oscillator. The solution method is the exact same. We assume only small deviations, which allows us to say sin(θ)≈θ
.
The pendulum will thus undergo simple harmonic motion, just like a regular simple pendulum. The solution to get to the period is straightforward now. It is given by (Thank you to Isaac Clark for pointing out that my previous expression didn’t have the right dimensionality since I wrote down the solution for the angular frequency.):
T=2π43Gπρ√=3πGρ−−−√
This period is, as you can see, independent of the mass of the pendulum, but also independent of the length of the pendulum cord.
PLZ FOLLOW ME
Explanation: