The end of a pendulum travels 50 cm on its first swing. Each swing after the first, it travels 99% as far as the preceding swing. How far will the pendulum tr

Question

The end of a pendulum travels 50 cm on its first swing.  Each swing after the first, it travels 99% as far as the preceding swing.  How far will the pendulum travel before it stops?

Answer 1

This is a geometric series problem, with r = 0.99 and a₁ = 50 cm.

 

The sum of a geometric series will give how far it travels:

S = a₁ / (1-r)

 

S = 50 / (1-0.99) = 50/0.01 = 5000 cm = 50 m

 

Hope this helps!

Answer 2

Rach,

 

You have a good answer from Cara.  I just want to expand on the explanation, in case you are not familiar with geometric series.  Are you familiar with them?  If not, look at it this way:

 

Swing 1:  50

Swing 2:  0.99(50)

Swing 3:  0.99[0.99(50] = (0.99)²(50)

Swing 4:  0.99[(0.99)²(50)] = (0.99)³(50)

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.

.

Swing n:  (0.99)^n-1(50)

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.

.

 

Do you see the pattern?  Now add all the swing distances and get

 

S = 50 + 50(0.99) + 50(0.99)² + 50(0.99)³ + ...

 

Let x = 0.99

 

S = 50(1 + x + x² + x³ + ...)

 

For |x| < 1,

 

1 + x + x² + x³ + ... = 1/(1  -x)

 

Then

 

S = 50/(1 - 0.99) cm,

 

as in Cara's answer.