Question Four
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the
gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces
it 10 o
from the equilibrium position, and releases it. Using a stopwatch, the student determines that the
period of the pendulum is 1.44 s. Determine the experimental value of the gravitational acceleration in
two decimal places and the angular speed of the pendulum.
Period=2PIsqrt(l/g) solve for g.
To determine the experimental value of the gravitational acceleration and the angular speed of the pendulum, we can use the formula for the period of a simple pendulum.
The formula for the period of a pendulum is given by:
T = 2π√(L/g)
Where:
T = Period of the pendulum
L = Length of the pendulum
g = Gravitational acceleration
In this case, the student has measured the length of the pendulum (L) as 0.510 meters and determined the period (T) as 1.44 seconds.
To find the experimental value of the gravitational acceleration (g), we can rearrange the formula as follows:
g = (4π²L) / T²
Substituting the known values gives:
g = (4π² * 0.510) / (1.44²)
Let's calculate it step by step:
Step 1: Calculate (1.44²)
1.44² = 2.0736
Step 2: Calculate (4π² * 0.510)
4π² = 39.478
39.478 * 0.510 = 20.19598
Step 3: Calculate (20.19598 / 2.0736)
20.19598 / 2.0736 = 9.7449
Therefore, the experimental value of the gravitational acceleration is approximately 9.74 m/s².
To find the angular speed of the pendulum, we can use the formula:
ω = 2π / T
Where:
ω = Angular speed of the pendulum
T = Period of the pendulum
Substituting the known value of T = 1.44 seconds, we can calculate:
ω = 2π / 1.44
Let's calculate it:
Step 1: Calculate (2π)
2π = 6.2832
Step 2: Calculate (6.2832 / 1.44)
6.2832 / 1.44 = 4.356
Therefore, the angular speed of the pendulum is approximately 4.36 rad/s.