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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. |
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A block is attached to a horizontal spring. The block is held so the spring is stretched and the block is released from rest, undergoing simple harmonic motion with a frequency of \( 2 \) \( \text{Hz} \). How long after release will the block first reach a point where it is momentarily at rest?
A cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
A pendulum has a period of \(2.0 \, \text{s}\) on Earth. What is its length?
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?
A spring launches a \(4 \, \text{kg}\) block across a frictionless horizontal surface. The block then ascends a \(30^\circ\) incline with a kinetic friction coefficient of \(\mu_k = 0.25\), stopping after \(55 \, \text{m}\) on the incline. If the spring constant is \(800 \, \text{N/m}\), find the initial compression of the spring. Disregard friction while in contact with the spring.
A linear spring of negligible mass requires a force of \( 18.0 \, \text{N} \) to cause its length to increase by \( 1.0 \, \text{cm} \). A sphere of mass \( 75.0 \, \text{g} \) is then attached to one end of the spring. The distance between the center of the sphere \( M \) and the other end \( P \) of the un-stretched spring is \( 25.0 \, \text{cm} \). Then the sphere begins rotating at constant speed in a horizontal circle around the center \( P \). The distance \( P \) and \( M \) increases to \( 26.5 \, \text{cm} \).
A pendulum consists of a ball of mass \( m \) suspended at the end of a massless cord of length \( L \). The pendulum is drawn aside through an angle of \( 60^\circ \) with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
A \( 240 \) \( \text{kg} \) block is dropped from \( 3.0 \) meters onto a spring, compresses the spring and comes to rest.
A box having a mass of \( 1.5 \) \( \text{kg} \) is accelerated across a table at \( 1.5 \) \( \text{m/s}^2 \). The coefficient of kinetic friction on the box is \( 0.3 \).
A \(10 \, \text{meter}\) long pendulum on the earth, is set into motion by releasing it from a maximum angle of less than \(10^\circ\) relative to the vertical. At what time \(t\) will the pendulum have fallen to a perfectly vertical orientation?
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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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