A pendulum bob swings 5.0 cm on its first oscillation. On each subsequent Oscillation the Bob travels 2/5 of the previous distance. Find the total distance the Bob travels before coming to a rest.
This is just a GP with
a = 5
r = 2/5
S∞ = a/(1-r) = 25/3
To find the total distance the bob travels before coming to a rest, we need to sum the distances traveled on each oscillation.
On the first oscillation, the bob swings 5.0 cm.
On the second oscillation, the bob travels (2/5) * 5.0 cm = 2.0 cm.
On the third oscillation, the bob travels (2/5) * 2.0 cm = 0.8 cm.
On the fourth oscillation, the bob travels (2/5) * 0.8 cm = 0.32 cm.
And so on...
We can represent this pattern as a geometric series with the first term (a) as 5.0 cm and the common ratio (r) as 2/5.
The sum of a geometric series can be found using the formula:
Sn = a * (1 - r^n) / (1 - r)
where Sn is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
Let's calculate the total distance traveled by the bob before coming to a rest:
Sn = 5.0 * (1 - (2/5)^n) / (1 - 2/5)
Since the bob comes to rest when the distance becomes negligible, we can consider the infinite sum by taking the limit as n approaches infinity:
Sn = 5.0 * (1 - 0) / (1 - 2/5)
Sn = 5.0 / (3/5)
Sn = 5.0 * (5/3)
Sn = 25/3
Sn ≈ 8.33 cm
Therefore, the total distance traveled by the bob before coming to a rest is approximately 8.33 cm.
To find the total distance the bob travels before coming to a rest, we need to sum up the distances traveled on each oscillation.
Let's break down the problem step by step:
First, we know that on the first oscillation, the bob swings 5.0 cm. Let's call this distance "d1".
On each subsequent oscillation, the bob travels 2/5 of the previous distance. This means that the distance traveled on the second oscillation is (2/5) times d1, the distance traveled on the third oscillation is (2/5) times (2/5) times d1, and so on.
We can set up a sequence of distances as follows:
d1 = 5.0 cm (distance on the first oscillation)
d2 = (2/5) * d1 (distance on the second oscillation)
d3 = (2/5) * d2 (distance on the third oscillation)
d4 = (2/5) * d3 (distance on the fourth oscillation)
...
From this pattern, we can see that the distance d2 is 2/5 of d1, d3 is 2/5 of d2, and so on.
To find the total distance the bob travels before coming to a rest, we need to sum up all these distances. We can use a geometric series to find the sum:
Total distance = d1 + d2 + d3 + d4 + ...
To express the sum in a simpler form, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
Where:
S = Total sum of the series
a = First term of the series
r = Common ratio of the series
In this case, a = d1 = 5.0 cm and r = 2/5.
Plugging these values into the formula, we get:
Total distance = 5.0 cm / (1 - 2/5)
Simplifying the denominator, we get:
Total distance = 5.0 cm / (3/5)
To divide by a fraction, we can multiply by its reciprocal:
Total distance = 5.0 cm * (5/3)
Calculating the result, we find:
Total distance = 25/3 cm ≈ 8.33 cm
Therefore, the total distance the bob travels before coming to a rest is approximately 8.33 cm.