1. Neha
Learning Object 2: 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