Simple and Physical Pendulums AP Physics C Mechanics MCQ – Exam Style Questions etc.
Simple and Physical Pendulums AP Physics C Mechanics MCQ
Unit 7: Oscillations
Weightage : 10-15%
Question
A pendulum and a spring both have a 1 kg sphere attached to their ends and have the same length when in equilibrium, as shown in the figure above. They both oscillate with the same period. If the 1 kg spheres are replaced with 2 kg spheres and the amplitudes of oscillation are unchanged, which of the following is true about the resultant period of oscillation for each?
Pendulum Spring
(A) Remains the same Decreases
(B) Remains the same Increases
(C) Remains the same Remains the same
(D) Increases Remains the same
(E) Increases Increases
Answer/Explanation
Ans:B
According to the equation for the period of a pendulum \(T_P=2\pi\sqrt{\frac{L}{g}}\), mass is not a factor; thus, the period of oscillation will remain the same. According to the equation for the period of a spring \(T_S=2\pi\sqrt{\frac{m}{k}}\), as the mass of the block increases, the period of oscillation increases.
Question(a)
A simple pendulum has a period of 2 s for small amplitude oscillations. 18. The length of the pendulum is most nearly
(A) 1/6 m
(B) 1/4 m
(C) 1/2 m
(D) 1m
(E) 2m
Answer/Explanation
Ans:D
Question(b)
Which of the following equations could represent the angle \(theta\) that the pendulum makes with the vertical as a function of time t ?
(A)\(\theta=\theta_{max}sin\frac{\pi }{2}t\)
(B)\(\theta=\theta_{max}sin\pi t\)
(C)\(\theta=\theta_{max}sin 2\pi t\)
(D)\(\theta=\theta_{max}sin 4\pi t\)
(E)\(\theta=\theta_{max}sin 8\pi t\)
Answer/Explanation
Ans:B
Question
A simple pendulum of length L, mass m, and amplitude A has a frequency of f on Earth. If this pendulum were moved to the Moon (which has \(\frac{1}{6}\) Earth’s gravity), what would be its new frequency?
(A) \((\frac{1}{6})^2 f\)
(B) \((\frac{1}{6})^{\frac{1}{2}} f\)
(C) \((6^2) f\)
(D) \((6)^{\frac{1}{2}}f\)
(E) f
Answer/Explanation
Ans: B
The formula for frequency of a pendulum is
\(f_{pendulum} = \frac{1}{2 \pi} \sqrt{(g/L)}\)
This means that the frequency is proportional to the square root of gravity. Therefore, since the moon’s gravity is \(\frac{1}{6}\) of Earth’s, the new frequency will be \((\frac{1}{6})^{\frac{1}{2}}\)
times the old frequency.
Question
A pendulum is launched into simple harmonic motion in two different ways, as shown above, from a point that is a height h above its lowest point. During both launches, the bob is given an initial speed of 3.0 m/s. On the first launch, the initial velocity of the bob is directed upward along the pendulum’s path, and on the second launch it is directed downward along the pendulum’s path. Which launch will cause the pendulum to swing with the larger amplitude?
(A) the first launch
(B) the second launch
(C) Both launches produce the same amplitude.
(D) The answer depends on the initial height h.
(E) The answer depends on the length of the supporting rope.
Answer/Explanation
Ans:
C—Consider the conservation of energy. At the launch point, the potential energy is the same regardless of launch direction. The kinetic energy is also the same because KE depends on speed alone and not direction. So, both balls have the same amount of kinetic energy to convert to potential energy, bringing the ball to the same height in every cycle.