| Derivation or Formula | Reasoning |
|---|---|
| \[ x(t) = A \cos\left(\omega t\right) \] | For simple harmonic motion the position is given by a cosine function with angular frequency \(\omega\). In a mass-spring system, \(\omega = \sqrt{\frac{k}{m}}\). |
| \[ x = A \cos\left(\sqrt{\frac{k}{m}}\, t\right) \] | This is option (a) which correctly substitutes \(\omega = \sqrt{\frac{k}{m}}\) into the general SHM equation. |
| \[ x = A \cos\left(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\, t\right) \] | Option (b) incorrectly scales the angular frequency by \(\frac{1}{2\pi}\); it does not match the standard angular frequency \(\sqrt{\frac{k}{m}}\). |
| \[ x = A \cos\left(2\pi\sqrt{\frac{m}{k}}\, t\right) \] | Option (c) uses \(\sqrt{\frac{m}{k}}\), the reciprocal of the correct value, and multiplies by \(2\pi\), leading to an incorrect angular frequency. |
| \[ x = A \cos\left(4\pi^2\sqrt{\frac{m}{k}}\, t\right) \] | Option (d) similarly uses \(\sqrt{\frac{m}{k}}\) with an extra factor of \(4\pi^2\) which deviates further from the correct form. |
| \[ \boxed{x = A \cos\left(\sqrt{\frac{k}{m}}\, t\right)} \] | Thus, only option (a) correctly represents the position function of a mass-spring system undergoing SHM. |
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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?
A pendulum has a period of \(2.0 \, \text{s}\) on Earth. What is its length?
Students attach a thin strip of metal to a table so that the strip is horizontal in relation to the ground. A section of the strip hangs off the edge of the table. A mass is secured to the end of the hanging section of the strip and is then displaced so that the mass-strip system oscillates, as shown in the figure. Students make various measurements of the net force \( F \) exerted on the mass as a result of the force due to gravity and the normal force from the strip, the vertical position \( y \) of the mass above and below its equilibrium position \( y \), and the period of oscillation \( T’ \) when the mass is displaced by different amplitudes \( A \). Which of the following explanations is correct about the evidence required to conclude that the mass undergoes simple harmonic motion?
What is the effect on the period of a pendulum if you double its length?
When the mass of a simple pendulum is tripled, the time required for one complete vibration
On Earth, a simple pendulum of length \(1.2 \, \text{meters}\), mass of \(3 \, \text{kg}\), and amplitude of \(10\) degrees oscillates back and forth. Calculate:
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?
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 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 \(1.0 \, \text{s}\) and \(1.1 \, \text{s}\) is most nearly
For an object oscillating in SHM, what is the relationship between its displacement, velocity, and acceleration graphs as a function of time?
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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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