Frequency and Period of SHM AP Physics C Mechanics MCQ – Exam Style Questions etc.
Frequency and Period of SHM AP Physics C Mechanics MCQ
Unit 7: Oscillations
Weightage : 10-15%
Question
A 0.50 kg object is attached to a vertical spring of constant k, as shown above. The object is pulled down and released. The object oscillates vertically. If up is the positive direction, the position x of the object as a function of time t is given by the formula \(x=\beta sin(\omega t+\Phi )\),where \(\beta =0.20\) , \(\omega =4.0rad/s\), and \(\varphi =\pi/3 rad\).
The period of the oscillation for the object is most nearly
(A) 4.0 s
(B) 3.1 s
(C) 2.0 s
(D) 1.6 s
(E) 1.1 s
Answer/Explanation
Ans:D
Question
A mass M suspended by a spring with force constant k has a period T when set into oscillation on Earth. Its period on Mars, whose mass is about \(\frac{1}{9}\) and radius\(\frac{1}{9}\) that of Earth, is most nearly
(A) \(\frac{1}{3}T \)
(B)\(\frac{2}{3}T \)
(C) \(T\)
(D) \(\frac{3}{2}T \)
(E) \(3T\)
Answer/Explanation
Ans:C
Question
A spring mass system is vibrating along a frictionless, horizontal floor. The spring constant is 8 N/m, the amplitude is 5 cm, and the period is 4 seconds.
In kg, the mass of the system is
(A) \(32 \pi ^2\)
(B) \(\frac{32}{\pi^2}\)
(C) \(\frac{16}{\pi}\)
(D) \(\frac{0.2}{\pi^2}\)
(E) \(\frac{20}{\pi ^2}\)
Answer/Explanation
Ans: B
The period for a spring-mass system is given by the equation \(T = 2 \pi \sqrt{\frac{m}{k}}\). Use that equation to solve for the mass, as shown below,
Question
A spring mass system is vibrating along a frictionless, horizontal floor. The spring constant is 8 N/m, the amplitude is 5 cm, and the period is 4 seconds.
Which of the following equations could represent the position of the mass from equilibrium x as a function of time t, where x is in meters and t is in seconds.
(A) x = 0.05 cos \(\pi t\)
(B) x = 0.05 cos \(2 \pi t\)
(C) x = 0.05 cos \(\frac{\pi}{2} t\)
(D) x = 8 cos \(\frac{\pi}{2} t\)
(E) x = 0.05 cos \(\frac{\pi}{4}t\)
Answer/Explanation
Ans: C
The equation for an object undergoing simple harmonic motion will be of the form \(x = A cos (\omega t + \phi )\), where A is the amplitude, \(\omega \) is the angular frequency and \(\phi \) is the phase constant. None of the solutions involve the phase constant, so we will ignore that part of the equation. The amplitude of the oscillation is 5 cm, which is 0.05 m. This eliminates (D). The angular frequency can be calculated by using the fact that the period is \(2 \pi\) divided by the angular frequency.