In Harmonic Motion how does the value of g affect the time period of a pendulum?
Period=2PI sqrt (length/g)
A 0.500kg particle moves in a circle of radius R = 0.150m at a constant speed. The time for 20 complete revolutions is 31.7s. what is the frequency of the circular motion?
In harmonic motion, the time period of a pendulum is affected by the value of acceleration due to gravity, denoted by 'g'. The time period of a pendulum refers to the time taken for one complete oscillation or swing.
To understand how the value of 'g' affects the time period, we need to consider the formula for the time period of a simple pendulum:
T = 2π√(L/g)
Where:
- T is the time period of the pendulum
- π is pi, approximately 3.14
- L is the length of the pendulum
- g is the acceleration due to gravity
From the formula, it is evident that 'g' is in the denominator under the square root (√). This means that as 'g' increases, the value of the denominator becomes larger, resulting in a smaller overall value inside the square root.
Since the value inside the square root becomes smaller, the entire expression becomes smaller. Hence, increasing the value of 'g' decreases the time period of the pendulum. On the other hand, decreasing 'g' would have the opposite effect and increase the time period.
This relationship shows that the time period of a pendulum is inversely proportional to the square root of the acceleration due to gravity.