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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T = 2\pi \sqrt{\frac{m}{k}}\] | Use the period formula for a mass–spring system, where \(T\) is the period, \(m\) the mass, and \(k\) the spring constant. |
| 2 | \[k = \frac{4\pi^2m}{T^2}\] | Solve for \(k\) by squaring the period equation and isolating \(k\). |
| 3 | \[k = \frac{4\pi^2 (5000)}{10^2} = \frac{4\pi^2 (5000)}{100} = 200\pi^2\] | Substitute \(m=5000\;\text{kg}\) and \(T=10\;\text{s}\) into the equation. |
| 4 | \[\boxed{k = 200\pi^2 \;\text{N/m}}\] | This is the final expression for the spring constant. |
Part (b): Equation of Motion
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[x(t) = A \cos(\omega t + \phi)\] | This is the standard form for simple harmonic motion, with amplitude \(A\), angular frequency \(\omega\), and phase \(\phi\). |
| 2 | \[A = 2,\quad \phi = 0\] | The elephant is pulled \(2\;\text{m}\) from equilibrium and released from rest, so the amplitude is \(2\;\text{m}\) and the initial phase is zero. |
| 3 | \[\omega = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5}\] | Calculate the angular frequency using the given period \(T=10\;\text{s}\). |
| 4 | \[x(t) = 2 \cos\Big(\frac{\pi}{5}t\Big)\] | Substitute \(A=2\), \(\omega=\pi/5\), and \(\phi=0\) into the standard equation. |
| 5 | \[\boxed{x(t) = 2 \cos\Big(\frac{\pi}{5}t\Big)}\] | This is the final equation of motion for the elephant on the spring. |
Part (c): Time to Travel from a Displacement of \(0.5\;\text{m}\) to \(1\;\text{m}\)
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[x(t) = 2 \cos\Big(\frac{\pi}{5}t\Big)\] | Recall the equation of motion from part (b). |
| 2 | \[2 \cos\Big(\frac{\pi}{5}t_1\Big) = 1\] | Set \(x(t_1)=1\;\text{m}\) to find the time \(t_1\) when the displacement is \(1\;\text{m}\). |
| 3 | \[\cos\Big(\frac{\pi}{5}t_1\Big) = \frac{1}{2}\] | Simplify the equation from step 2. |
| 4 | \[\frac{\pi}{5}t_1 = \cos^{-1}\Big(\frac{1}{2}\Big) = \frac{\pi}{3}\] | Use the inverse cosine; note that \(\cos^{-1}(1/2) = \pi/3\) within the relevant interval. |
| 5 | \[t_1 = \frac{5}{\pi}\cdot \frac{\pi}{3} = \frac{5}{3}\] | Solve for \(t_1\) by isolating it. |
| 6 | \[2 \cos\Big(\frac{\pi}{5}t_2\Big) = 0.5\] | Set \(x(t_2)=0.5\;\text{m}\) to determine the time \(t_2\) when the displacement is \(0.5\;\text{m}\). |
| 7 | \[\cos\Big(\frac{\pi}{5}t_2\Big) = 0.25\] | Simplify the equation from step 6. |
| 8 | \[\frac{\pi}{5}t_2 = \cos^{-1}(0.25)\] | Express \(t_2\) in terms of the inverse cosine. |
| 9 | \[t_2 = \frac{5}{\pi}\cos^{-1}(0.25)\] | Solve for \(t_2\) by isolating it. |
| 10 | \[\Delta t = \Big|t_2 – t_1\Big| = \frac{5}{\pi}\Big|\cos^{-1}(0.25) – \frac{\pi}{3}\Big|\] | The time interval required to travel between the two displacements is the difference between \(t_2\) and \(t_1\). The absolute value ensures a positive time difference regardless of the order of passage. |
| 11 | \[\boxed{\Delta t = \frac{5}{\pi}\Big(\cos^{-1}(0.25) – \frac{\pi}{3}\Big) \approx 0.43\;\text{s}}\] | This is the final expression and its approximate numerical value for the time interval. |
Just ask: "Help me solve this problem."
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 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 launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.
A [katex] 2 \, \text{kg}[/katex] mass is attached to a spring with spring constant [katex] k = 100 \, \text{N/m}[/katex] and negligible mass.
A 0.2 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.4 J. The kinetic energy as a function of position presented by the graph.
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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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