| 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. |
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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 \( t = \dfrac{T}{8} \)?
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 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
What is the relationship between the period \( T \) and frequency \( f \) of an object in simple harmonic motion?
An experimenter has a simple pendulum of length \( L \) and a mass–spring system with mass \( m \) and spring constant \( k \). Both are found to have the same period of oscillation \( T \) on Earth. If both systems are taken to the Moon, where the acceleration due to gravity is approximately \( \frac{1}{6} g \) of Earth, what will happen to their periods?
A \(20 \, \text{g}\) piece of clay moving at a speed of \(50 \, \text{m/s}\) strikes a \(500 \, \text{g}\) pendulum bob at rest. The length of a string is \(0.8 \, \text{m}\). After the collision, the clay-bob system starts to oscillate as a simple pendulum.
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?
A block with a mass of \( 4 \) \( \text{kg} \) is attached to a spring on the wall that oscillates back and forth with a frequency of \( 4 \) \( \text{Hz} \) and an amplitude of \( 3 \) \( \text{m} \). What would the frequency be if the block were replaced by one with one‑fourth the mass and the amplitude of the block is increased to \( 9 \) \( \text{m} \)?
What is the defining characteristic of the restoring force that causes an object to undergo Simple Harmonic Motion (SHM)?
A \( 2 \, \text{kg}\) mass is attached to a spring with spring constant \( k = 100 \, \text{N/m}\) and negligible mass.
\(2.86\,\text{rad/s}\)
\(2.20\,\text{s}\)
\(0.60\,\text{m/s}\)
\(1.71\,\text{m/s^{2}}\)
\(2.5\times 10^{2}\,\text{m/s^{2}}\)
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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] |
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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