1) The document provides information about a simple pendulum experiment with a basketball hanging from a string. It asks the learner to calculate various properties of the pendulum's motion. 2) The learner is asked to find the net force, tension, restoring force, acceleration, angular frequency, and displacement of the pendulum at different points in its swing. 3) Conceptual questions ask how increasing the string length or mass would affect the angular frequency of the pendulum's oscillation.

1. Neha
Learning Object 1: Simple Pendulum Challenge
Bill has a toy basketball with length 1.55m
and a mass of 125g hanging off the wall of his
room.
Show your work:
Q1: Find the net force at point P1 if the tension
acting on the string is 0.5N.
Q2: Billy swings the ball. What is the tension at
point P2? (∅ = 43)
Q3: What is the restoring force at P2 pulling
back the ball to equilibrium?
Q4: Calculate the acceleration of the tangential component at P2.
Q5: Find the angular frequency of this oscillation.
Q6: What is the displacement at point P2?
Conceptual Questions:
Q7: If the length of the string is increased to 3.0m, would the angular
frequency increase or decrease?
Q8: If the mass of the string is increased to 200g, would the angular
frequency increase or decrease?

2. Bonus:
What is the angle at ∅2?
Answers:
Q1: Since this is a mass hanging off of the string we have two forces to
take into account, Fg and Ft.
Solving for Fg:
Fg=mg
(125g/1000)(9.80m/s^2)
=1.23N
Solving for Fnet:
Fnet=Fg-Ft
Fnet=1.2325N-0.5N
Fnet=0.7N
Q2: Since the mass is now at an angle, we can solve for x and y
components of forces for that angle. The tension in the string is equal to
the force acting in the y direction (for every force there is an equal and
opposite force!).
Solving for T:
T=mgCos∅
T=(.125kg)(9.80m/s^2)(Cos43)
T=0.90N
Q3: It is the tangential component that provides the restoring force that
brings the ball back to equilibrium. Therefore, we must find the force
acting in the x direction.
Solving for Fr:

3. Fr=mgSin∅
Fr=(.125kg)(9.80m/s^2)(Sin43)
Fr=0.84N
Since this force is in the opposite direction, Fr=-0.84N
Q4: If we have the length of the string, we can use the angular frequency
formula.
W(angular frequency)=√(
𝑔
𝐿
)
W=√
9.80𝑚/𝑠^2
1.55𝑚
W=2.51 Hz
Q6: The displacement at P2 depends on the length of the string and the
angle at which the ball is at.
180 degrees= 𝜋 radians
therefore,
43 degrees=0.750 radians
S=L∅
S=(1.55m)(0.750 radians)
S=1.16m
Q7: Since the equation of angular frequency is√(
𝑔
𝐿
) , increasing the
length of the string will decrease the angular frequency.
Q8: There is no change. Angular frequency does not rely on mass, but
on length!

4. Bonus: At P1, the angle of the ball with respect to the wall is 90
degrees. If ∅1=43 degrees, then:
∅2=90-43=47 degrees