Question

. Anna wishes to make a simple pendulum that serves as a timing device. She plans to make it such that
its period is 1.00 second on Earth. What length must the pendulum have?
11. Anna takes her pendulum to planet B that has twice the mass of Earth with the same radius as Earth.
What is the period of her pendulum on planet B?
12. Next, Anna travels to planet C that has half the mass of Earth with a radius twice that of Earth. What is
the period on planet C?
13. Finally, Anna takes her pendulum to planet D. On planet D, Anna’s pendulum has a period of 1.4
seconds. Calculate the acceleration due to gravity of planet D.
14. Somewhere on a distant planet, a simple pendulum is pulled away from the equilibrium point and
released. The pendulum comes back to the point of release exactly 2.4 seconds after the release. If the
length of the pendulum is 1.3 m, what is the acceleration due to gravity on the planet?

Answer 1

Answer:

24.8 cm (0.248 m)

Explanation:

Use the equation T = 2•π•(L/g).5

Insert 1.00 s in for T and 9.8 m/s2 in for g. Give attention to your algebra:

Square both sides of the equation to remove the radical.

The equation becomes L = T2•g/(4•π2).

The length comes out to be 0.24824 m. Round to three significant digits.

Answer 2

The formula for a simple pendulum of length "L" having a harmonic motion in gravity "g" has a Time period of oscillation "T" as :

                          $T = 2$π$\sqrt{L/g}$

10. The length of the pendulum for a 1 sec time period on Earth is 0.25m or 25 cm.

Given :  $T = 1 sec, g= 10 ms^{-2}$

The length of the pendulum "L" according to the formula is :

            $L = T^{2}*g/(4*3.14*3.14)$

                $= (1*1*10)/(4*3.14*3.14)$

                $= 10/39.44$

                $= 0.253 m$

11. The Time period of the pendulum on planet B will be 0.71 seconds.

The formula for gravity "g" on a planet with mass "M" and radius "R" has a Gravitational constant "G".

                         $g = G*M/R^{2}$

Planet B has the mass twice and the radius same, so the gravity also becomes twice as Mass is in direct proportion to gravity.

                         $T_{new} = 2*3.14\sqrt{L/g_{new} }$

                                   $= 2*3.14\sqrt{L/(2*g) }$

                                   $= (1/\sqrt{2}) * 2*3.14\sqrt{L/g}$

                                   $= (1/\sqrt{2}) * T$

                                   $=1/\sqrt{2}$

                                   $=0.71 sec$

12. The Time period of the pendulum on planet C will be 2.83 seconds.

Planet C has a mass half and a radius twice of Earth, so the gravity also becomes one-eighth of that of Earth.

                         $T_{new} = 2*3.14\sqrt{L*R_{new} ^{2} /G * M_{new} }$

                                   $= 2*3.14\sqrt{L*(2R) ^{2} /G * (M/2) }$

                                   $= \sqrt{8} * 2*3.14\sqrt{L/g}$

                                   $= \sqrt{8} * T$

                                   $=2.83sec$

13. The gravity of planet D is $5.02 ms^{-2}$.

Given :  $T = 1.4 sec, L = 0.25m$

The gravity of the planet according to the formula is :

            $g = (4*3.14*3.14*L)/T^{2}$

                $= (4*3.14*3.14*0.25)/(1.4*1.4)$

                $= 9.85/1.96$

                $= 5.02 ms^{-2}$

14. The gravity of the distant planet E is $8.9 ms^{-2}$.

Given :  $T = 2.4 sec, L = 1.3m$

The gravity of the planet according to the formula is :

            $g = (4*3.14*3.14*L)/T^{2}$

                $= (4*3.14*3.14*1.3)/(2.4*2.4)$

                $= 51.27/5.76$

                $= 8.9 ms^{-2}$

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