AP Physics C: Mechanics practice test 6.

Congratulations - you have completed .

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

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?

$x(t)=2.5 \ \cos(ωt+10°)$ | |

$x(t)=5 \ \tan(ωt+10°)$ | |

$x(t)=5 \ \sin(ωt)$ | |

$x(t)=2.5 \ \cos(ωt+5°)$ | |

$x(t)=5 \ \sin(ωt+10°)$ |

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

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?

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 $-Aω \ \sin \ ωt$

$x=A \cos ωt$

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 $t=0$?

$0.62$ | |

$0.43$ | |

$0.70$ | |

$0.54$ | |

$0.38$ |

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 $

$=y(t=0)$

$=7 \cos(\dfrac{π}{6}$

$=6.06$

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?

$0.44 \ m/s$ | |

$0.28 \ m/s$ | |

$0.35 \ m/s$ | |

$0.11 \ m/s$ | |

$0.23 \ m/s$ |

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,

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

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

$=\dfrac{1}{2}×20×0.04^2$

$=10×0.0016$

$=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.

$0 \ sec$ | |

$π/2 \sec $ | |

$π \ sec$ | |

$3π/2 \ sec$ | |

$2π \ sec$ |

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 $t=π \ s$

(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 $t = π/ω$.

$x_o ω^2 \sinϕ$ | |

$x_o ω^2$ | |

$x_o ω^2 \cosϕ$ | |

$x_o ω^2 \sin \dfrac{ϕ}{2}$ | |

$x_o ω^2 \tanϕ$ |

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 \sinϕ $

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

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

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 \sinϕ $

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?

$\sqrt{\dfrac{9k}{m}}$ | |

$\sqrt{\dfrac{3k}{m}}$ | |

$\sqrt{\dfrac{k}{3m}}$ | |

$\sqrt{\dfrac{k}{m}}$ | |

$\sqrt{\dfrac{k}{9m}}$ |

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

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?

The potential energy of the system is approximately a parabola for $x_l ≤ x ≤ x_r $ | |

The system could perform simple harmonic oscillations for $x ≥ x_r$ | |

The kinetic energy of the system is maximum when it is at $x_r$ or $x_l$ | |

The potential energy of the system has an unstable equilibrium between $x_l$ and $x_r $ | |

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

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?

$f=\dfrac{1}{2π} \sqrt{\dfrac{j^2}{m}}$ | |

$=4π \sqrt{\dfrac{j^2}{m}}$ | |

$f=\dfrac{1}{2π} \dfrac{m}{j}$ | |

$f=\dfrac{1}{2π} \dfrac{ \sqrt{m}}{j}$ | |

$f=\dfrac{1}{π} \dfrac{ \sqrt{m}}{j}$ |

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}}$

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

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?

$0.45 \ m/s$ | |

$0.39 \ m/s$ | |

$0.36 \ m/s$ | |

$0.28 \ m/s$ | |

$0.22 \ m/s$ |

Question 10 Explanation:

Let initial compression be $x_1$ and equilibrium compression be $x_2$

$x_2=\dfrac{mg \ \sinθ}{k}$

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

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 $ $=0.245 \ J$

$h=0.04×\sin30°=0.02 \ m$

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

$→v=0.28 \ m/s$

$x_2=\dfrac{mg \ \sinθ}{k}$

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

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 $ $=0.245 \ J$

$h=0.04×\sin30°=0.02 \ m$

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 $m/s^2$?

$4.4 \ m/s^2$ | |

$7.3 \ m/s^2$ | |

$6.5 \ m/s^2$ | |

$9.8 \ m/s^2$ | |

$5.0 \ m/s^2$ |

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 $=\dfrac{4π^2}{g}=7.9$

$g=\dfrac{4π^2}{7.9}≈5 \ m/s^2$

$→ω^2=\dfrac{g}{l}$ 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 $t=\dfrac{4T}{3}$?

$3.55 \ \dfrac{v_o \ T}{2π}$ | |

$4.87 \ \dfrac{v_o \ T}{2π}$ | |

$5.50 \ \dfrac{v_o \ T}{2π}$ | |

$4.00 \ \dfrac{v_o \ T}{2π}$ | |

$6.25 \ \dfrac{v_o \ T}{2π}$ |

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

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

$=\int -v_o \ \sin\dfrac{2πt}{T} dt$

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

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

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

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 12 questions to complete.

List |

**Next Practice Test:
Unit 7: Gravitation >>**

AP Physics C Mechanics Main Menu >>