What determines the period of a pendulum?
2 answers:
The length of the pendulum and the acceleration of gravity determine the period of a pendulum.
<u>Explanation: </u>
We all are familiar with the pendulum and its motion. The way it swings from its rest position to the left and right ends with specific amplitude.
A pendulum can be simply defined as an object hung of a static point with a bob connected to it through a string.
When the bob is raised to the either direction from its rest position, it swings back and forth because of the acceleration (g). A complete back and forth swing of a bob from its rest position in the pendulum is counted as one complete period of a pendulum.
This primarily depends on length and acceleration. For simple pendulums, formula to evaluate its period is,
Where,
L- length of the string from the static suspension point to the bob in the pendulum
g- Acceleration due to gravity = 9.8
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The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7º
given parameters
- the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
- the movement times t = 1.0s, 2.0s and 3.0 s
to find
a) position
b) acceleration
c) launch angle
Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration
b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.
v_y = v_{oy} - g t
where v and v({oy} are the velocities of the body, g the acceleration of the planet's gravity and t the time
0 = v_{oy} - gt
g = v_{oy} / t
from the graph we observe that the highest point occurs for t = 2.0 s
g = 1.2 / 2.0
g = 0.6 m / s²
a) The position is requested for several times
X axis
in this axis there is no acceleration so we can use the uniform motion relationships
vₓ = x / t
x = vₓ t
where x is the position, vx is the velocity and t is the time
we calculate for the time
t = 0.0 s
x₀ = 0
t = 2.0 s
x₂ = 1.8 2
x₂ = 3.6 m
t = 3.0 s
x₃ = 1.8 3
x₃ = 5.4 m
Y axis
In this axis there is the acceleration of the planet, let us use for the position the relation
y = v_{oy} t - ½ g t²
t = 0.0 s
y₀ = 0
y₀ = 0 m
t = 2.0 s
y₂ = 1.2 2 - ½ 0.6 2²
y₂ = 1.2 m
t = 3.0 s
y₃ = 1.2 3 - ½ 0.6 3²
y₃ = 0.9 m
c) the launch angle use the trigonometry relation
tan θ =
θ = tan⁻¹
θ = tan⁻¹
θ = 33.7º
measured counterclockwise from the positive side of the x-axis
With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7ºto)
learn more about projectile launch here: