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(a) Calculation of the angular frequency (\(\omega\)) of the oscillation:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\omega = \sqrt{\frac{g}{L}}\) | The angular frequency formula for a simple pendulum, where \(g\) is the acceleration due to gravity \(9.8 \, \text{m/s}^2\) and \(L\) is the length of the pendulum. |
| 2 | \(\omega = \sqrt{\frac{9.8}{1.2}}\) | Substitute \(g = 9.8 \, \text{m/s}^2\) and \(L = 1.2 \, \text{m}\) into the formula. |
| 3 | \(\omega = \sqrt{8.17}\) | Calculate the division inside the square root. |
| 4 | \(\omega \approx 2.86 \, \text{rad/s}\) | Calculate the square root to find the angular frequency. |
(b) Calculation of the period (\(T\)) of oscillation:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(T = \frac{2\pi}{\omega}\) | The period of oscillation is the reciprocal of the angular frequency multiplied by \(2\pi\). |
| 2 | \(T = \frac{2\pi}{2.86}\) | Substitute the angular frequency value found earlier. |
| 3 | \(T \approx 2.20 \, \text{s}\) | Calculate the period. |
(c) Calculation of the maximum speed of the pendulum:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_{\text{max}} = \omega A\) | Maximum speed is the product of angular frequency and amplitude. |
| 2 | \(A = 1.2 \cdot \sin(10^\circ)\) | Convert amplitude from degrees to a distance using the pendulum length (1.2 m) and the sine of 10 degrees. |
| 3 | \(A \approx 0.208\, \text{m}\) | Calculate the amplitude in meters. |
| 4 | \(v_{\text{max}} = 2.86 \cdot 0.208\) | Substitute the values of \( \omega \) and \( A \) to find maximum velocity. |
| 5 | \(v_{\text{max}} \approx 0.595 \, \text{m/s}\) | Compute the maximum speed. |
(d) Calculation of the maximum acceleration of the pendulum:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(a_{\text{max}} = \omega^2 A\) | Maximum acceleration is the product of the square of angular frequency and amplitude. |
| 2 | \(a_{\text{max}} = (2.86)^2 \times 0.208\) | Substitute the values of \( \omega \) and \( A \). |
| 3 | \(a_{\text{max}} \approx 1.70 \, \text{m/s}^2\) | Calculate the maximum acceleration. |
(e) Determine the acceleration due to gravity on the Exoplanet:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(f = \frac{1}{T}\) | Frequency is the reciprocal of the period. |
| 2 | \(T = \frac{1}{2.3}\) | Substitute the frequency of oscillation (2.3 Hz). |
| 3 | \(T \approx 0.435 \, \text{s}\) | Compute the period of oscillation on the Exoplanet. |
| 4 | \(T = 2\pi \sqrt{\frac{L}{g_{\text{exo}}}}\) | Use the period expression in terms of gravity and pendulum length to solve for gravity on the Exoplanet. |
| 5 | \(\frac{0.435}{2\pi} = \sqrt{\frac{1.2}{g_{\text{exo}}}}\) | Rearrange to express \(\frac{T}{2\pi}\) in terms of known values. |
| 6 | \(g_{\text{exo}} = \frac{1.2}{(0.435/(2\pi))^2}\) | Solve for the gravitational acceleration. |
| 7 | \(g_{\text{exo}} \approx 90 \, \text{m/s}^2\) | Calculate to find the acceleration due to gravity on the Exoplanet. |
Just ask: "Help me solve this problem."
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A [katex] 2 \, \text{kg}[/katex] mass is attached to a spring with spring constant [katex] k = 100 \, \text{N/m}[/katex] and negligible mass.
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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 s and 1.1 s is most nearly
A block attached to spring demonstrates simple harmonic motion about its equilibrium position with amplitude [katex] A [/katex] and angular frequency [katex] \omega [/katex]. What is the maximum magnitude of the block’s velocity?
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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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