Respuesta :
The period of a simple pendulum depends only on the length of the pendulum and the gravitational acceleration:
[tex]T=2 \pi \sqrt{ \frac{L}{g} } [/tex]
where L is the pendulum length and g the gravitational acceleration.
The problem says that Maya and the swing form a simple pendulum, so we can use this formula to calculate the period of Maya's motion, using the length of the swing (L=1.8 m):
[tex]T=2 \pi \sqrt{ \frac{1.8 m}{9.81 m/s^2} }=2.69 s [/tex]
[tex]T=2 \pi \sqrt{ \frac{L}{g} } [/tex]
where L is the pendulum length and g the gravitational acceleration.
The problem says that Maya and the swing form a simple pendulum, so we can use this formula to calculate the period of Maya's motion, using the length of the swing (L=1.8 m):
[tex]T=2 \pi \sqrt{ \frac{1.8 m}{9.81 m/s^2} }=2.69 s [/tex]
Answer:
The period is 2.69 seconds.
Explanation:
The period of the pendulum can be defined as the time taken by the pendulum to motion back from the equilibrium position and then forth the equilibrium position and back to equilibrium position.
Briefly it is time for one complete oscillation.
It is dependent on the length of pendulum. The mass of pendulum has no effect on the period.
T = 2π√L/g
Here g is gravitational acceleration, its value is constant on the earth and is 9.8 m/s2.
By putting values in formula:
T = 2(3.14)√1.8/9.8
T = 2.69 s
Hence the period is 2.69 seconds.