A pendulum bob swings 6.0 cm on its first oscillation. On each subsequent oscillation the bob travels 1/3 of the previous distance. Find the total distance the bob travels before coming to rest.
6 + 6/3 + 6/3^2 + 6/3^3 + ...
we have a GS, where a = 6 and r = 1/3
sum(infinite number of terms) = a/(1-r)
= 6/(1-1/3)
= 6/(2/3)
= 6*3/2 = 9
check:
sum(1) = 6
sum(2) = 6 + (1/3)(6) = 8
sum(3) = 6 + 6/3 + 6/9 = 8 2/3
sum(4) = ....... = 8 8/9
sum(5) = .... 8 26/27
etc
Ah, the swinging pendulum bob, quite the traveler! Let's calculate its total distance before it decides to take a rest and go on a swinging holiday.
On the first oscillation, the bob swings 6.0 cm. Now, on each subsequent oscillation, it travels 1/3 of the previous distance. So, for the second oscillation, it would be 6.0 cm * 1/3 = 2.0 cm.
For the third oscillation, it would be 2.0 cm * 1/3 = 0.67 cm (approximately).
We can do this for all the subsequent oscillations until the bob comes to rest. However, eventually, the distance becomes very small and negligible, so we'll stop after a few oscillations.
Let's add up all these distances to find the total distance! *summoning calculator*
6.0 cm + 2.0 cm + 0.67 cm = 8.67 cm (approximately).
So, the pendulum bob will travel a total distance of approximately 8.67 cm before coming to a rest. Quite the journey for a swinging bob, wouldn't you say? Keep swinging, Bob!
To find the total distance the pendulum bob travels before coming to rest, we need to sum up the distances of all its oscillations.
Let's start by finding the distances traveled during the first few oscillations:
First oscillation: 6.0 cm
Second oscillation: 1/3 * 6.0 cm = 2.0 cm
Third oscillation: 1/3 * 2.0 cm = 0.67 cm (rounded to two decimal places)
We can continue this pattern to find the distances for subsequent oscillations:
Fourth oscillation: 1/3 * 0.67 cm = 0.22 cm (rounded to two decimal places)
Fifth oscillation: 1/3 * 0.22 cm = 0.07 cm (rounded to two decimal places)
...
Nth oscillation: 1/3 * previous distance
Now we can see that the distances keep getting smaller and smaller. To simplify the calculation, we can use a geometric series to find the sum of the distances:
S = a / (1 - r)
Where:
- "S" is the sum of the distances
- "a" is the first term of the series (6.0 cm)
- "r" is the common ratio (1/3)
Plugging in the values, we get:
S = 6.0 cm / (1 - 1/3)
S = 6.0 cm / (2/3)
S = 6.0 cm * (3/2)
S = 9.0 cm
Therefore, the total distance the bob travels before coming to rest is 9.0 cm.
To find the total distance the bob travels before coming to rest, we need to sum up the distances traveled on each oscillation.
On the first oscillation, the bob swings 6.0 cm.
On the second oscillation, the bob travels 1/3 of the distance traveled on the first oscillation, which is (1/3) * 6.0 cm = 2.0 cm.
On the third oscillation, the bob travels 1/3 of the distance traveled on the second oscillation, which is (1/3) * 2.0 cm = 0.67 cm (rounded to two decimal places).
We can continue this pattern and add up the distances traveled on each oscillation until the bob comes to rest.
Let's calculate it step by step:
1st oscillation: 6.0 cm
2nd oscillation: (1/3) * 6.0 cm = 2.0 cm
3rd oscillation: (1/3) * 2.0 cm = 0.67 cm
4th oscillation: (1/3) * 0.67 cm ≈ 0.22 cm
5th oscillation: (1/3) * 0.22 cm ≈ 0.07 cm
...
As the bob swings for a larger number of oscillations, the distance traveled on each subsequent oscillation becomes smaller and smaller.
We can see that the distances traveled form a decreasing geometric sequence. The common ratio between each term is 1/3.
To find the sum of an infinite geometric series, where the absolute value of the common ratio is less than 1, we can use the formula:
Sum = a / (1 - r),
where "a" represents the first term and "r" represents the common ratio.
In this case, the first term "a" is 6.0 cm and the common ratio "r" is 1/3.
Sum = 6.0 cm / (1 - 1/3)
Sum = 6.0 cm / (2/3)
Sum = 6.0 cm * (3/2)
Sum = 9.0 cm
Therefore, the total distance the bob travels before coming to rest is 9.0 cm.