# 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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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