AP Physics C Mechanics: Unit 6 Practice Test — Oscillations

AP Physics C: Mechanics practice test 6.

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Question 1

A simple pendulum oscillates with an amplitude of $5 \ cm$ and has a phase angle of $10°$. Which of the below options possibly shows the equation of motion for the simple pendulum?

A
$x(t)=2.5 \ \cos⁡(ωt+10°)$
B
$x(t)=5 \ \tan⁡(ωt+10°)$
C
$x(t)=5 \ \sin⁡(ωt)$
D
$x(t)=2.5 \ \cos⁡(ωt+5°)$
E
$x(t)=5 \ \sin⁡(ωt+10°)$
Question 1 Explanation: 
Option A $→$ It is incorrect because, amplitude should be $5 \ cm$ and not $2.5 \ cm$

Option B $→$ It is incorrect because, $\tan \ x$ should be a $\cos⁡ \ x$ or $\sin⁡ \ x$

Option C $→$ It is incorrect because, phase angle should be $10°$ and not $0°$

Option D $→$ It is incorrect because, amplitude should be $5 \ cm$ and not $2.5 \ cm$

Option E $→$ It is correct because, $x(t)$ is of the form $A \sin⁡ (ωt+ϕ)$ where $A$ is the amplitude and $ϕ$ is the phase angle
Question 2

A spring-mass system is set into motion by stretching the spring to the right of its equilibrium position and then released.

Which of the below graphs shows the correct velocity-time graphs for the system after it is released?

A
B
C
D
E
Question 2 Explanation: 
After stretching the spring-mass system to the right and then released the position time function should be of the form

$x=A \cos ⁡ωt$

And so, the velocity time function should be of the form

$v=-Aω \sin⁡ ωt$

Option C represents the function $-Aω \ \sin \ ⁡ωt$
Question 3

A simple harmonic oscillator moves according to the equation $y(t)=7 \cos⁡\left{dfrac{πt}{4} + \dfrac{π}{6}\right)$.

When will the oscillator be $2$ units away from the starting position after it starts oscillating from $t=0$?

A
$0.62$
B
$0.43$
C
$0.70$
D
$0.54$
E
$0.38$
Question 3 Explanation: 
Starting position of the oscillator

$=y(t=0)$

$=7 \cos⁡(\dfrac{π}{6}$

$=6.06$

Position of oscillator when it is $2$ units away from the starting position

$=6.06 - 2$

$=4.06$

$7 \ cos⁡ \left(\dfrac{πt}{4} + \dfrac{π}{6}\right)=4.06$

Solving for t,

$\dfrac{πt}{4} + \dfrac{π}{6}= \cos^{-1}⁡0.58$

$t=0.54 $
Question 4

A block of mass $2.5 \ kg$ is attached to a spring of constant $20 \ N/m$. The system is then set into Simple Harmonic Motion by compressing the spring by $4 \ cm$. What is the maximum velocity that the block can have during its motion?

A
$0.44 \ m/s$
B
$0.28 \ m/s$
C
$0.35 \ m/s$
D
$0.11 \ m/s$
E
$0.23 \ m/s$
Question 4 Explanation: 
Potential energy stored in the spring

$=\dfrac{1}{2} kA^2$

$=\dfrac{1}{2}×20×0.04^2$

$=10×0.0016$

$=0.016 \ J $

Maximum kinetic energy of the block

$=\dfrac{1}{2} mv_{max}^2 $

$=0.016 \ J $

Solving for $v_{max}$ gives,

$v_{max}=\sqrt{√2×\dfrac{0.016}{2.5}}=0.11 \ m/s$
Question 5

Look at the position-time graph for a simple pendulum

Find the value of time $t$ when the velocity of the simple pendulum reverses its direction for the $1st$ time.

A
$0 \ sec$
B
$π/2 \sec $
C
$π \ sec$
D
$3π/2 \ sec$
E
$2π \ sec$
Question 5 Explanation: 
From time $t=0 \ s$ to $t=π \ s$ the pendulum moved from $x=-4$ to $x=4$

(in the $+x$ direction)

And from time $t=π \ s$ to $t=2π \ s$ the pendulum moved from $x=4$ to $x=-4$

(in the $-x$ direction)

Thus, the velocity of the pendulum changed at time $t=π \ s$
Question 6

The equation of motion for a simple harmonic oscillator is given by the relationship,

$x(t)=x_o \sin⁡(ωt+ϕ)$

Derive an expression for the time-dependent acceleration $a(t)$ of the simple harmonic oscillator and thus work out the acceleration at time $t = π/ω$.

A
$x_o ω^2 \sin⁡ϕ$
B
$x_o ω^2$
C
$x_o ω^2 \cos⁡ϕ$
D
$x_o ω^2 \sin⁡ \dfrac{ϕ}{2}$
E
$x_o ω^2 \tan⁡ϕ$
Question 6 Explanation: 
$x(t)=x_o \ \sin⁡ (ωt+ϕ)$

$v(t)=\dfrac{dx}{dt}=x_o ω \ \cos⁡ (ωt+ϕ)$

$a(t)=\dfrac{d^2 x}{dt^2}=-x_o ω^2 \ \sin (ωt+ϕ)$

For $t=\dfrac{π}{ω}$

$a\left(\dfrac{π}{ω}\right)=-x_o ω^2 \sin⁡(ω π/ω+ϕ)$

             $=-x_o ω^2 \sin⁡(π+ϕ)$

             $=x_o ω^2 \sin⁡ϕ $
Question 7

Look at the below spring-mass system.

If the spring is displaced slightly to the right, as shown above, then what is the angular frequency $ω$ of the system?

A
$\sqrt{\dfrac{9k}{m}}$
B
$\sqrt{\dfrac{3k}{m}}$
C
$\sqrt{\dfrac{k}{3m}}$
D
$\sqrt{\dfrac{k}{m}}$
E
$\sqrt{\dfrac{k}{9m}}$
Question 7 Explanation: 
Let the displacement of the mass be $x$

Net force acting on the mass $=-3kx$

Thus, $ma=-3kx$

$m \dfrac{d^2 x}{dt^2}=-3kx$

On rearranging, $\dfrac{d^2 x}{dt^2} + \dfrac{3k}{m} x=0$

Thus, $ω^2=\dfrac{3k}{m}→ω=\sqrt{\dfrac{3k}{m}}$
Question 8

The graph represents the potential energy PE of a system as a function of its position $x$.

When the system is displaced slightly from its equilibrium position $(x_l ≤ x ≤ x_r)$, it oscillates approximately as a simple harmonic oscillator. Which of the below statements is true regarding the system?

A
The potential energy of the system is approximately a parabola for $x_l ≤ x ≤ x_r $
B
The system could perform simple harmonic oscillations for $x ≥ x_r$
C
The kinetic energy of the system is maximum when it is at $x_r$ or $x_l$
D
The potential energy of the system has an unstable equilibrium between $x_l$ and $x_r $
E
The kinetic energy of the system would be independent of the potential energy
Question 8 Explanation: 
Option A $→$ True, as the system performs simple harmonic oscillations for $x∈(x_l,x_r)$

Option B $→$ False, as the potential energy of the system deviates from a parabola for large $x$

Option C $→$ False, as the system is performing simple harmonic motion, the minimum kinetic energy is at its end points ($x_r$ or $x_l$)

Option D $→$ False, the system has a stable equilibrium for $x∈(x_l,x_r)$

Option E $→$ False, the total energy of a simple harmonic oscillator is a constant and so the kinetic energy and potential energy are dependent on each other
Question 9

The differential equation of a rotating system is given as:

$m \dfrac{d^2 θ}{dt^2} = -j^2 \ θ$

Where m is the mass of the system, $θ$ is the angle of rotation, and $j$ is a positive constant. What is the frequency of the rotating system?

A
$f=\dfrac{1}{2π} \sqrt{\dfrac{j^2}{m}}$
B
$=4π \sqrt{\dfrac{j^2}{m}}$
C
$f=\dfrac{1}{2π} \dfrac{m}{j}$
D
$f=\dfrac{1}{2π} \dfrac{ \sqrt{m}}{j}$
E
$f=\dfrac{1}{π} \dfrac{ \sqrt{m}}{j}$
Question 9 Explanation: 
$m \dfrac{d^2 \ θ}{dt^2}=-j^2 \ θ$

$\dfrac{d^2 \ θ}{dt^2}=-\dfrac{j^2}{m} \ θ$

Or$ \dfrac{d^2 \ θ}{dt^2} + \dfrac{j^2}{m} \ θ=0 $

$\dfrac{d^2 θ}{dt^2}+ω^2 \ θ=0$

Here $ω=\sqrt{\dfrac{j^2}{m}}=2πf$

Thus, $ f=\dfrac{1}{2π} \ \sqrt{ \dfrac{j^2}{m}}$
Question 10

A spring-block system is shown below.

When this system is in equilibrium, the spring is compressed from its natural length. The spring is then compressed by another $4 \ cm$ and then released. What is the velocity of the block when it passes through its equilibrium position?

A
$0.45 \ m/s$
B
$0.39 \ m/s$
C
$0.36 \ m/s$
D
$0.28 \ m/s$
E
$0.22 \ m/s$
Question 10 Explanation: 
Let initial compression be $x_1$ and equilibrium compression be $x_2$

$x_2=\dfrac{mg \ \sin⁡θ}{k}$

$x_2=\dfrac{0.5×10× \sin30°}{25}=0.1 \ m$

Thus, $x_1=0.1+0.04=0.14 \ m$

Initial energy $=\dfrac{1}{2} \ kx_1^2$

$=\dfrac{1}{2}×25×0.14^2 $ $=0.245 \ J$

$h=0.04×\sin⁡30°=0.02 \ m$

Final energy $=mgh+\dfrac{1}{2} \ kx_2^2+\dfrac{1}{2} \ mv^2$

$=0.5×10×0.02+\dfrac{1}{2}×25×0.1^2+\dfrac{1}{2}×0.5×v^2 $ $=0.225+0.25v^2$

Thus, $v^2=\dfrac{0.245-0.225}{0.25=0.08}$

$→v=0.28 \ m/s$
Question 11

A simple pendulum satisfies the differential equation $\dfrac{d^2 x}{dt^2} = -\dfrac{g}{l} \ x$. The graph below shows the variation of the time period squared with the length of the pendulum.

What is the value of $g$ in $m/s^2$?

A
$4.4 \ m/s^2$
B
$7.3 \ m/s^2$
C
$6.5 \ m/s^2$
D
$9.8 \ m/s^2$
E
$5.0 \ m/s^2$
Question 11 Explanation: 
$\dfrac{d^2 x}{dt^2}=-\dfrac{g}{l} x$

$→ω^2=\dfrac{g}{l}$ OR $ω=\sqrt{\dfrac{g}{l}}$

Using $ω=\dfrac{2π}{T}$ and solving for $T^2$ gives,

$T^2=\dfrac{4π^2}{g} \ l$

Slope of the line $=\dfrac{4π^2}{g}=7.9$

$g=\dfrac{4π^2}{7.9}≈5 \ m/s^2$
Question 12

If $v(t)=-v_o \sin⁡ \dfrac{2πt}{T}$ for a simple harmonic oscillator starting from $x(t=0) = \dfrac{v_o \ T}{2π}$ (where v_o>0, and $T$ is the time period of the oscillator), then what is the total distance traveled by the oscillator from time $t=0$ to $t=\dfrac{4T}{3}$?

A
$3.55 \ \dfrac{v_o \ T}{2π}$
B
$4.87 \ \dfrac{v_o \ T}{2π}$
C
$5.50 \ \dfrac{v_o \ T}{2π}$
D
$4.00 \ \dfrac{v_o \ T}{2π}$
E
$6.25 \ \dfrac{v_o \ T}{2π}$
Question 12 Explanation: 
$x=\int v(t) \ dt$

$=\int -v_o \ \sin⁡\dfrac{2πt}{T} dt$

$=C+\dfrac{v_o T}{2π} \ \cos⁡ \dfrac{2πt}{T}$

For $t=0, x=\dfrac{v_o \ T}{2π}$

$C+\dfrac{v_o \ T}{2π}= \dfrac{v_o \ T}{2π}→C=0$

$x=\dfrac{v_o T}{2π} \ \cos \dfrac{2πt}{T}$

In one complete period $T$, the distance traveled by the oscillator is $4 \dfrac{v_o \ T}{2π}$

While the distance traveled in time $\dfrac{T}{3}$ is given by,

$|x(T/3)-x(0) |= | \dfrac{v_o \ T}{2π} \ \cos⁡ \left( \dfrac{2π}{T} \dfrac{T}{3}\right) - \dfrac{v_o \ T}{2π}|=$

$1.5 \dfrac{v_o \ T}{2π}$

Total distance traveled in time

$\dfrac{4T}{3}=4 \dfrac{v_o \ T}{2π}+1.5 \dfrac{v_o \ T}{2π}=5.5 \dfrac{v_o \ T}{2π}$
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