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AP Physics C Mechanics Energy of Simple Harmonic Oscillators MCQ

Energy of Simple Harmonic Oscillators AP  Physics C Mechanics MCQ – Exam Style Questions etc.

Energy of Simple Harmonic Oscillators AP  Physics C Mechanics MCQ

Unit 7: Oscillations

Weightage : 20-15%

AP Physics C Mechanics Exam Style Questions – All Topics

Questions (a)-(c)

A block of mass m is on a rough horizontal surface and is attached to a spring with spring constant k. The coefficient of kinetic friction between the surface and the block is m. When the block is at
position x = 0, the spring is at its unstretched length. The block is pulled to position x = +x 0 , as shown above, and released from rest. The block then travels to the left and passes through x = 0 before coming momentarily to rest at position\( x = – x_0 /2\) .

Question(a)

Which of the following is a correct expression for the kinetic energy of the block as it first travels through position x = 0?
(A) 0
(B) \(kx^2_0\)
(C)\( kx^2_0/2-\mu mgx_0\)
(D) \(kx^2_0/2-\mu mgx_0/2\)
(E)\(kx^2_0/2-\mu mgx_0\)

Answer/Explanation

Ans:C

The kinetic energy at x = 0 is equal to the difference of the initial elastic potential energy of the spring-block system and the energy dissipated by friction. Substituting into an energy equation

\(U_{Si}+fd=K_f\)

yeilds \(K_f=\frac{1}{2}kA^2-fd=\frac{1}{2}kx^2_0-\mu mdx_o\)

Question(b)

Which of the following is a correct expression for the coefficient of kinetic friction \(\mu\)?

(A)\(\frac{kx_0}{4mg}\)
(B)\(\frac{kx_0}{2mg}\)
(C)\(\frac{3kx_0}{4mg}\)
(D)\(\frac{kx_0}{mg}\)
(E)\(\frac{2kx_0}{mg}\)

Answer/Explanation

Ans:A

Setting the energy dissipated by friction equal to the difference in the elastic potential energy of the spring-block system

Question(c)

 Which of the following is a differential equation that could be used to solve for the block’s position x as a function of time t when it is moving to the left?

(A)\(m\frac{d^2x}{dt^2}=kx+\mu mg\)

(B)\(m\frac{d^2x}{dt^2}=kx-\mu mg\)

(C)\(m\frac{d^2x}{dt^2}=-kx+\mu mg\)

(D)\(m\frac{d^2x}{dt^2}=-kx-\mu mg\)

(E)\(m\frac{d^2x}{dt^2}=kx\)

Answer/Explanation

Ans:C

Substituting into a Newton’s second law equation yields

\(F_{net}=F_S+f=ma\)

\(m\frac{\partial^2 x}{\partial t^2}=-kx+\mu mg\)

Questions(a) , (b)

A block on a horizontal frictionless plane is attached to a spring, as shown above. The block oscillates along the x-axis with simple harmonic motion of amplitude A.

Question(a)

Which of the following statements about the block is correct?
(A) At x = 0, its velocity is zero.
(B) At x = 0, its acceleration is at a maximum.
(C) At x = A, its displacement is at a maximum.
(D) At x = A, its velocity is at a maximum.
(E) At x = A, its acceleration is zero.

Answer/Explanation

Ans:C

Solution: Basic fact about SHM. Amplitude is max displacement

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