| 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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A student is designing an experiment to find the spring constant \( k \) of a spring using only a set of known masses and a stopwatch. Which procedure would work?
Block \( 1 \) of mass \( m_1 \) and Block \( 2 \) of mass \( m_2 = 2 m_1 \) are each attached to identical horizontal springs. Each block is displaced from equilibrium by an unknown amount and the blocks are released from rest simultaneously, undergoing simple harmonic motion. A student claims that Block \( 1 \) will make its first return to its equilibrium position before Block \( 2 \) first returns to its equilibrium position. Is this claim correct? Why or why not?
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:
The total energy of a system in SHM is given by \( E \). At what displacement from equilibrium is the kinetic energy equal to the potential energy?
What is the defining characteristic of the restoring force that causes an object to undergo Simple Harmonic Motion (SHM)?
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.
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 linear spring of force constant \( k \) is used in a physics lab experiment. A block of mass \( m \) is attached to the spring and the resulting frequency, \( f \), of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If \( f^{2} \) is plotted versus \( \frac{1}{m} \), the graph will be a straight line with slope
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 \(t = 3.5\) \(\text{s}\)?
A block attached to spring demonstrates simple harmonic motion about its equilibrium position with amplitude \( A \) and angular frequency \( \omega \). What is the maximum magnitude of the block’s velocity?
\(2.86\,\text{rad/s}\)
\(2.20\,\text{s}\)
\(0.60\,\text{m/s}\)
\(1.71\,\text{m/s^{2}}\)
\(2.5\times 10^{2}\,\text{m/s^{2}}\)
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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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