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| Derivation/Formula | Reasoning |
|---|---|
| \[ \omega = \sqrt{\frac{k}{m}} \] | This is the formula for the angular frequency of a mass-spring system, where \(k=20.0\,\text{N/m}\) and \(m=1.5\,\text{kg}\). |
| \[ \omega = \sqrt{\frac{20.0}{1.5}} \approx 3.65\,\text{rad/s} \] | Substitute the given values to calculate \(\omega\). |
| \[ f = \frac{\omega}{2\pi} \approx \frac{3.65}{6.28} \approx 0.582\,\text{Hz} \] | Convert the angular frequency to the ordinary frequency using \(f=\omega/(2\pi)\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[ v_{\text{max}} = A\,\omega \] | The maximum speed in simple harmonic motion is the product of the amplitude \(A\) and the angular frequency \(\omega\). |
| \[ v_{\text{max}} = 0.10\,\text{m} \times 3.65\,\text{rad/s} \approx 0.365\,\text{m/s} \] | Substitute the amplitude \(A=0.10\,\text{m}\) and the computed \(\omega\) into the formula. |
| \[ \text{Occurs at } x=0 \] | The maximum speed occurs at the equilibrium position where the displacement is zero. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ a_{\text{max}} = \omega^2\,A \] | The maximum acceleration in simple harmonic motion is given by \(a_{\text{max}}=\omega^2 A\). |
| \[ a_{\text{max}} = (3.65\,\text{rad/s})^2 \times 0.10\,\text{m} \approx 1.33\,\text{m/s}^2 \] | Substitute \(\omega \approx 3.65\,\text{rad/s}\) and \(A = 0.10\,\text{m}\) into the equation. |
| \[ \text{Occurs at } x = \pm 0.10\,\text{m} \] | The magnitude of acceleration is maximum at the extreme positions (\(x=\pm A\)) of the oscillation. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ E = \frac{1}{2}\,k\,A^2 \] | The total mechanical energy in a mass-spring system is stored as potential energy in the spring at maximum displacement. |
| \[ E = \frac{1}{2} \times 20.0\,\text{N/m} \times (0.10\,\text{m})^2 \] | Substitute the given values \(k=20.0\,\text{N/m}\) and \(A=0.10\,\text{m}\) into the energy formula. |
| \[ E = 0.1\,\text{J} \] | Simplify the expression to obtain the total energy of the system. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ x(t) = A\,\cos(\omega t + \phi) \] | This is the general solution for the displacement in simple harmonic motion, where \(\phi\) is the phase constant. |
| \[ x(0) = A\,\cos(\phi) = 0.10\,\text{m} \] | At \(t=0\), the mass is released from rest at \(x=0.10\,\text{m}\), which implies \(\phi = 0\) because \(\cos(0)=1\). |
| \[ x(t) = 0.10\,\text{m}\,\cos\Big(\sqrt{\frac{20.0}{1.5}}\,t\Big) \] | Substitute \(A=0.10\,\text{m}\), \(\omega=\sqrt{\frac{20.0}{1.5}}\), and \(\phi=0\) into the general solution to obtain the displacement as a function of time. |
Just ask: "Help me solve this problem."
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
A 0.5 kg cart, on a frictionless 2 m long table, is being pulled by a 0.1 kg mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls 0.5 m, calculate the speed of the cart on the table. Use ONLY forces and energy.
An apple is released from rest 500 m above the ground. Due to the combined forces of air resistance and gravity, it has a speed of 40 m/s when it reaches the ground. What percentage of the initial mechanical energy of the apple-Earth system was dissipated due to air resistance? Take the potential energy of the apple-Earth system to be zero when the apple reaches the ground.
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 kickball is rolled by the pitcher at a speed of 10 m/s and it is kicked by another student. The kickball deforms a little during the kick, and then rebounds with a velocity of 15 m/s as its shape restores to a perfect sphere. Select all that must be true about the kickball and the kicking foot system.
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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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