Listen to the rhythm of the falling weights
Introduction
You can space weights along a string so that they make a regular rhythm of beats when they strike the ground.
Material
Some string, such as parachute cord (1/8 to 1/4 inch diameter, 3 to 6 mm) two pieces about 9 feet, or 3 m, long.
10 Weights to clip on the string, such as 1/2 ounce (15 gram) fishing weights, or binder clips.
A ruler
Optional, a cookie sheet
Assembly
On string number one, clip the weights at the following distances from one end, "the bottom end".
50 cm,100,cm,150,cm, 200 cm, 250 cm
On string number two clip the weights at
10 cm, 40,cm, 90 cm, 160 cm, 250 cm
Optional place the cookie sheet on the floor, drop the weights on the cookie sheet.
To Do and Notice
Hold string one by the weight at 250 cm so that the bottom of the string barely touches the floor. Notice that there is no weight at the bottom end near the floor.
You may have to stand on a step stool to hold the string high enough off the floor.
Drop the weights and listen to the rhythm of the falling weights.
Notice how the rhythm goes faster and faster as the weights fall.
Now repeat the drop with string two. Hold it by the weight at 250 cm from the bottom so that the bottom of the string barely brushes the floor or the cookie sheet. Notice that there is no weight at the bottom.
Drop the string and listen for the rhythm.
Notice that the rhythm has a constant beat. There is a constant time interval between the weights as they hit the floor.
What's Going On?
The weights fall under gravity, they accelerate downward.
Each weight falls a distance proportional to the square of the time that it falls.
Thus equally spaced weights will hit with a shorter and shorter time interval between them.
In order to hit after equal time intervals the weights must be spaced with a distance that increases proportional to a square.
Notice that the distances of the weights are spaced proportional to the squares of the number of each weight, 1, 4, 9, 16 etc.
So string two has weights spaced to fall at equal time intervals under free fall while string one is spaced at equal distances.
Math Root
For string two:
An object falls from rest with constant acceleration, g then the distance it travels at a time t is
d = 1/2 gt^{2}
So if a weight falls a distance d it takes a time t = (2*d/g)^{0.5}
For a weight to fall for exactly twice this time it the travels
d = 1/2 g t^{2} = 1/2 g (2*(2*d/g^{)0.5})^{2} = 1/2g 4 2d/g = 4d
show that after three time intervals the weight falls a distance of 9d.
and so on.
For String one:
The time it takes the lowest weight at d = 50 cm to fall to the ground is
t1 = (2*d/g)^{0.5}
The time it takes for weight number 2 at twice the distance from the ground 2d to hit the ground is:
t2 = (2*2d/g)^{0.5} = 2^{0.5} * (2*d/g)^{0.5} = 2^{0.5} t1
Show that the time it takes weight 3 to hit the ground is
t3 = 3^{0.5} t1
and for weight n is n^{0.5} t1
Going Further
You can record the sound of the falling weights and analyze them on a computer.
Sound1 is the sound oof thhe equally spaced weights recorded oon a computer and turned into an mp3 file. You can definitely hear that thhe last interval is shoorter than the first.
Sound2 is tthe recording of thhe string two with weights spaced along distannces pproportional to the square of the weight number.
Scientific Explorations with Paul Doherty 

24 October 2004 