finding the acceleration of a simple pendulum
g= 4pi^2 x length/T^2
T^2= 2.28
length= 70cm
Acceleration in what direction? horizontal? vertical? centripetal? angular?
Acceleration at what time? What is the amplitude of the oscillation? That affects the answer.
The horizontal (x) acceleration for small displacements varies with time as
ax = -(amplitude)*w^2*sin wt
where
w = sqrt(g/L)
its a simple pendulum brass ball.
hung from a string swinging with no applied pressure. the pendulum was drawn into the air 20 cm from the center. and swings to the other side. same distance 20 cm
So the amplitude is 20 cm. You can use the formula I gave for the x acceleration, but you never said which acceleration you want.
To find the acceleration of a simple pendulum, we can use the equation:
acceleration (a) = g * (length / T^2)
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2 or 32.2 ft/s^2).
- length is the length of the pendulum in meters or feet.
- T is the period of the pendulum in seconds.
In your case, you have given the length of the pendulum (70 cm) and the value of T^2 (2.28). However, before we proceed with the calculation, we need to convert the length to meters.
Given:
length = 70 cm = 0.7 m
T^2 = 2.28
Now, substitute these values into the equation:
a = (4 * pi^2 * length) / T^2
a = (4 * 3.14^2 * 0.7) / 2.28
Evaluating the expression:
a ≈ 2.836 m/s^2
Therefore, the acceleration of the simple pendulum is approximately 2.836 m/s^2.