What is the period, in seconds, of a simple pendulum of length 3 meters?
Use the gravitational constant g = 9.8 m/s2 and round your answer to two decimal places
Please help! A step by step would with the answer would be helpful!
the answer is 3.48
D man is correct, it is 3.48.
3.48 just like what D man said
To find the period of a simple pendulum, you can use the formula:
T = 2π√(L/g)
Where:
T = period of the pendulum
π = 3.14 (approximation for pi)
L = length of the pendulum
g = acceleration due to gravity
In this case, you are given:
L = 3 meters
g = 9.8 m/s^2
Now we can plug in these values into the formula:
T = 2π√(3/9.8)
First, divide 3 by 9.8:
T = 2π√(0.3061)
Next, take the square root of 0.3061:
T = 2π×0.5539
Now, multiply 2π by 0.5539:
T = 3.48 seconds (rounded to two decimal places)
Therefore, the period of a simple pendulum with a length of 3 meters is approximately 3.48 seconds.
surely you have the formula for the period of a pendulum of length L (and small amplitude):
T = 2π √(L/g)
Now just plug in your numbers.