Suppose that a tunnel is dug through the center of the earth. You stand on a tower 100 m tall and drop a 3-kg rock into the tunnel. What is the frequency of the resulting harmonic motion?

The frequency of the resulting harmonic motion is 0.000219 Hz

Step-by-step explanation:

We are going to calculate the time it takes for one single wave ocillation.

Frequency and the time taken to finish a single wave oscillation are inversely proportional. The formula for calculating frequency when given the time taken to complete a wave cycle is written as: f = 1 / T

In this formula, f represents frequency and T represents the time period or amount of time required to complete a single wave oscillation.

I consider the initial speed to be zero, because it is of no significance compared with the free fall into the earth, through the earth and back again.

Given from Wikipedia:

The diameter of the earth is 1.2742 * 10⁴ km which is 1.27 * 10⁷ m

2 times the radius = diameter, so the radius of the earth = (1.27 * 10⁷ m) /2 = 6.4 * 10⁶ m

radius earth = r

r = 6.4 * 10⁶ m

Now imagine the tunnel and the free fall.

1. Initially the rock has no speed.

2. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

3. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed and it has travelled the distance r !

4. After this moment, the Rock will be slowed down because of the negative accelleration...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely, and again passing the distance r.

5. Now at the other end of the earth there is the same initial situation as described at point 1, only the Rock has travelled the distance equal to the diameter of the earth, (exactly 2 times r).

So basically, the samething happens once more, only this time it starts exactly from the other end of the earth...

6. Initially the rock has no speed.

7. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

8. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed.

9. By now the Rock will be slowed down because of the negative accelleration... It is moving towards the initial starting point...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely.

10. Now finally the Rock is exactly at the starting position.

In reality there will have been some loss of speed due to friction, so the Rock will be slightly lower then the 100 m above the ground.

let's calculate the time it takes to free fall for the distance r.

initial speed =0 and after 6.4 * 10⁶ m it's speed will be maximum. We need to find out how much time passes before that distance is passed.

r = v*t + 0.5*a*t²

r = 0 + 0.5*a*t²

0.5*a*t² = r

t² = r / ( 0.5 * a )

t² = 6.4 *10⁶ / ( 0.5 * 9.8 )

t² = 1.306 * 10 ⁶

t = 1142.86 s

Now please confirm that in order for the Rock to move back to the initial starting point it has to travel 4 times as much time. It has to travel r to centre of the earth then another r to travel to to the other side of the earth, and back again. So indeed 4 times r.

The time it will take must be the same as 4 * 1142.86 s

now this is the time of one single wave ocillation.

Since T = 4571.43 s

f = 1 / 4571.43

f = 0.00021874993164 Hz

The frequency of the resulting harmonic motion is 2.19 *10-4

The frequency of the resulting harmonic motion is 0.000219 Hz

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah