| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = A\cos(\omega t)\] | This is the equation for the displacement of a block in simple harmonic motion, where \(A\) is the amplitude. |
| 2 | \[v = \frac{d(\Delta x)}{dt} = -A\omega\sin(\omega t)\] | The velocity is the time derivative (slope) of the displacement equation. The factor \(\omega\) comes from differentiating \(\cos(\omega t)\). |
| 3 | \[|v|_{\text{max}} = A\omega\] | The maximum magnitude of the velocity occurs when \(|\sin(\omega t)| = 1\), so \(|v|_{\text{max}} = A\omega\). |
| 4 | \[\boxed{A\omega}\] | Final answer indicating that the maximum velocity is \(A\omega\), which corresponds to answer choice (a). |
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An object undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
The graph represents the position \( x \) as a function of time \( t \) for an object undergoing simple harmonic motion. Which of the following equations could represent the position \( x \) as a function of time \( t \)?
A block is attached to a horizontal spring. The block is held so the spring is stretched and the block is released from rest, undergoing simple harmonic motion with a frequency of \( 2 \) \( \text{Hz} \). How long after release will the block first reach a point where it is momentarily at rest?
A block of mass \( 0.5 \) \( \text{kg} \) is attached to a horizontal spring with a spring constant of \( 150 \) \( \text{N/m} \). The block is released from rest at position \( x = 0.05 \) \( \text{m} \), as shown, and undergoes simple harmonic motion, reaching a maximum position of \( x = 0.1 \) \( \text{m} \). The speed of the block when it passes through position \( x = 0.09 \) \( \text{m} \) is most nearly
A bullet (mass: \(0.05 \, \text{kg}\)) is fired horizontally (\(v = 200 \, \text{m/s}\)) at a block (mass: \(1.3 \, \text{kg}\)) initially at rest on a frictionless surface. The block is attached to a spring (\(k = 2500 \, \text{N/m}\)). The bullet becomes embedded. Calculate:
A simple pendulum oscillates with amplitude \(A\) and period \(T\), as represented on the graph above. Which option best represents the magnitude of the pendulum’s velocity \(v\) and acceleration \(a\) at time \(\frac{T}{2}\)?
A Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
A \(0.50 \, \text{kg}\) mass is attached to a spring constant \(20 \, \text{N/m}\) along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of \(1.5 \, \text{m/s}\) at the equilibrium position. What is the total energy of the system?
At time \( t = 0 \), an object is released from rest at position \( x = +x_{\text{max}} \) and undergoes simple harmonic motion along the \( x \)-axis about the equilibrium position of \( x = 0 \). The period of oscillation of the object is \( T \). Which of the following expressions is equal to the object’s position at time \( t = \dfrac{T}{8} \)?
A student sets an object attached to a spring into oscillatory motion and uses a motion detector to record the velocity of the object as a function of time. The total change in the object’s speed between \(1.0 \, \text{s}\) and \(1.1 \, \text{s}\) is most nearly
The maximum magnitude of the block’s velocity is \(\boxed{A\omega}\).
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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