0 attempts
0% avg
UBQ Credits
| 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."
What force is necessary to stretch an ideal spring with a spring constant of \( 120 \) \( \text{N/m} \) by \( 30 \) \( \text{cm} \)?
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 \(0.50 \, \text{kg}\) mass is attached to a spring constant \(20 \, \text{N/m}\) along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of \(1.5 \, \text{m/s}\) at the equilibrium position. What is the total energy of the system?
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 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
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
