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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 Christmas ornament made from a thin hollow glass sphere hangs from a thin wire of negligible mass. It is observed to oscillates with a frequency of \( 2.50 \) \( \text{Hz} \) in a city where \( g = 9.80 \) \( \text{m/s}^2 \). What is the radius of the ornament? The moment of inertia of the ornament is given by \( I = \frac{5}{3} mr^2 \).
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 mass–spring system is oscillating in simple harmonic motion. At the exact moment the mass passes through its equilibrium position, which of the following statements is true?
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 0.2 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.
A block of mass \( 0.5 \) \( \text{kg} \) is attached to a horizontal spring with a spring constant of \( 150 \) \( \text{N/m} \). The block is released from rest at position \( x = 0.05 \) \( \text{m} \), as shown, and undergoes simple harmonic motion, reaching a maximum position of \( x = 0.1 \) \( \text{m} \). The speed of the block when it passes through position \( x = 0.09 \) \( \text{m} \) is most nearly
A \( 240 \) \( \text{kg} \) block is dropped from \( 3.0 \) meters onto a spring, compresses the spring and comes to rest.
A block is attached to a horizontal spring and is initially at rest at the equilibrium position \( x = 0 \), as shown in Figure \( 1 \). The block is then moved to position \( x = -A \), as shown in Figure \( 2 \), and released from rest, undergoing simple harmonic motion. At the instant the block reaches position \( x = +A \), another identical block is dropped onto and sticks to the block, as shown in Figure \( 3 \). The two–block–spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy \( E_{\text{tot},2} \) of the two–block–spring system after the collision to the total mechanical energy \( E_{\text{tot},1} \) of the one–block–spring system before the collision?
A \( 50 \) \( \text{g} \) ice cube can slide up and down a frictionless \( 30^{\circ}\) slope. At the bottom, a spring with spring constant \( 25 \) \( \text{N/m} \) is compressed \( 10 \) \( \text{cm} \) and is used to launch the ice cube up the slope. How high does it go above its starting point? Express your answer with the appropriate units.
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?
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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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