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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\omega = \sqrt{\dfrac{g}{L}}\] | For small angles, a simple pendulum performs simple harmonic motion with angular frequency given by \(\sqrt{g/L}\). |
| 2 | \[\omega = \sqrt{\dfrac{9.81\,\text{m/s}^2}{1.2\,\text{m}}}\] | Substitute \(g = 9.81\,\text{m/s}^2\) and \(L = 1.2\,\text{m}\). |
| 3 | \[\boxed{\omega = 2.86\,\text{rad/s}}\] | Evaluate the square root. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T = \dfrac{2\pi}{\omega}\] | The period of simple harmonic motion is the reciprocal of the frequency: \(T = 2\pi/\omega\). |
| 2 | \[T = \dfrac{2\pi}{2.86}\] | Insert the value of \(\omega\) from part (a). |
| 3 | \[\boxed{T = 2.20\,\text{s}}\] | Compute the quotient. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\theta_{\max} = 10^{\circ} \times \dfrac{\pi}{180^{\circ}}\] | Convert the amplitude from degrees to radians: \(10^{\circ}=0.1745\,\text{rad}\). |
| 2 | \[\Delta x_{\max} = L\,\theta_{\max}\] | Arc length for small angles: \(\Delta x = L\theta\). |
| 3 | \[v_{\max} = \omega\,\Delta x_{\max}\] | For SHM, maximum speed equals \(\omega\) times maximum displacement. |
| 4 | \[v_{\max} = 2.86\,(1.2)(0.1745)\] | Insert \(\omega\), \(L\), and \(\theta_{\max}\). |
| 5 | \[\boxed{v_{\max} = 0.60\,\text{m/s}}\] | Multiply to find the speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a_{\max} = \omega^{2}\,\Delta x_{\max}\] | In SHM, maximum acceleration equals \(\omega^{2}\) times maximum displacement. |
| 2 | \[a_{\max} = (2.86)^2\,(0.209)\] | Use \(\Delta x_{\max}=0.209\,\text{m}\) from part (c). |
| 3 | \[\boxed{a_{\max} = 1.71\,\text{m/s}^{2}}\] | Compute the product. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\omega = 2\pi f\] | Angular frequency relates to frequency: \(\omega = 2\pi f\). |
| 2 | \[g’ = L\,\omega^{2}\] | For a pendulum, \(\omega^{2} = g’/L\Rightarrow g’ = L\omega^{2}\). |
| 3 | \[\omega = 2\pi(2.3)\] | Insert \(f = 2.3\,\text{Hz}\). |
| 4 | \[g’ = 1.2\,[2\pi(2.3)]^{2}\] | Substitute \(L = 1.2\,\text{m}\) and the expression for \(\omega\). |
| 5 | \[\boxed{g’ = 2.5 \times 10^{2}\,\text{m/s}^{2}}\] | Evaluate to obtain the exoplanet’s gravitational acceleration. |
Just ask: "Help me solve this problem."
A student uses a pendulum to determine the acceleration due to gravity, \( g \). They measure the pendulum’s length \( L \) and its period \( T \). Which equation should they use to calculate \( g \)?
A pendulum with a period of \( 1 \) \( \text{s} \) on Earth, where the acceleration due to gravity is \( g \), is taken to another planet, where its period is \( 2 \) \( \text{s} \). The acceleration due to gravity on the other planet is most nearly
A 0.4 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.2 J. The potential energy as a function of position presented by the graph.
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 \( I = \frac{5}{3} mr^2 \).
Three pendulums are set in motion, oscillating through small amplitudes. Each has the same mass. Rank the period of the pendulums from shortest to longest.
\(2.86\,\text{rad/s}\)
\(2.20\,\text{s}\)
\(0.60\,\text{m/s}\)
\(1.71\,\text{m/s^{2}}\)
\(2.5\times 10^{2}\,\text{m/s^{2}}\)
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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] |
General Metric Conversion Chart
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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