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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."
A \( 0.30 \text{-kg} \) mass is suspended on a spring. In equilibrium the mass stretches the spring \( 2.0 \) \( \text{cm} \) downward. The mass is then pulled an additional distance of \( 1.0 \) \( \text{cm} \) down and released from rest. Write down its equation of motion.
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
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?
A pendulum has a period of 2.0 s on Earth. What is its length?
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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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