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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."
An object in simple harmonic motion obeys the following position versus time equation: \( y = (0.50 \text{ m}) \sin \left( \frac{\pi}{2} t \right) \). What is the amplitude of vibration?
A \(10 \, \text{meter}\) long pendulum on the earth, is set into motion by releasing it from a maximum angle of less than \(10^\circ\) relative to the vertical. At what time \(t\) will the pendulum have fallen to a perfectly vertical orientation?
For an object oscillating in SHM, what is the relationship between its displacement, velocity, and acceleration graphs as a function of time?
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 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 \(\frac{T}{2}\)?
\(2.86\,\text{rad/s}\)
\(2.20\,\text{s}\)
\(0.60\,\text{m/s}\)
\(1.71\,\text{m/s^{2}}\)
\(2.5\times 10^{2}\,\text{m/s^{2}}\)
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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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