The period of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
– \( T \) is the period,
– \( L \) is the length of the pendulum,
– \( g \) is the acceleration due to gravity.
Let’s compare the periods of the three pendulums based on their lengths.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( T_1 = 2\pi \sqrt{\frac{L}{g}} \) | The period of pendulum 1, with length \( L \). |
| 2 | \( T_2 = 2\pi \sqrt{\frac{L}{g}} \) | The period of pendulum 2, also with length \( L \). |
| 3 | \( T_3 = 2\pi \sqrt{\frac{2L}{g}} \) | The period of pendulum 3, with length \( 2L \). |
| 4 | \( T_3 = 2\pi \sqrt{\frac{2L}{g}} = 2\pi \sqrt{2} \sqrt{\frac{L}{g}} = \sqrt{2} \cdot T_1 \) | Since \( \sqrt{2} > 1 \), the period \( T_3 \) is greater than \( T_1 \) and \( T_2 \). |
| 5 | Rank: \( T_1 = T_2 < T_3 \) | Pendulums 1 and 2 have equal periods, and pendulum 3 has the longest period. |
The correct answer is (b) \( T_1 = T_2 > T_3 \).
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A block with a mass of \( 4 \) \( \text{kg} \) is attached to a spring on the wall that oscillates back and forth with a frequency of \( 4 \) \( \text{Hz} \) and an amplitude of \( 3 \) \( \text{m} \). What would the frequency be if the block were replaced by one with one‑fourth the mass and the amplitude of the block is increased to \( 9 \) \( \text{m} \)?
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.
Block \( 1 \) of mass \( m_1 \) and Block \( 2 \) of mass \( m_2 = 2 m_1 \) are each attached to identical horizontal springs. Each block is displaced from equilibrium by an unknown amount and the blocks are released from rest simultaneously, undergoing simple harmonic motion. A student claims that Block \( 1 \) will make its first return to its equilibrium position before Block \( 2 \) first returns to its equilibrium position. Is this claim correct? Why or why not?
What is the relationship between the period \( T \) and frequency \( f \) of an object in 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:
On Earth, a simple pendulum of length \(1.2 \, \text{meters}\), mass of \(3 \, \text{kg}\), and amplitude of \(10\) degrees oscillates back and forth. Calculate:
A simple pendulum oscillates with amplitude \(A\) and period \(T\), as represented on the graph above. Which option best represents the magnitude of the pendulum’s velocity \(v\) and acceleration \(a\) at time \(\frac{T}{2}\)?
A block attached to a spring demonstrates simple harmonic motion about its equilibrium position with amplitude \( A \) and angular frequency \( \omega \). What is the maximum magnitude of the block’s velocity?
For an object oscillating in SHM, what is the relationship between its displacement, velocity, and acceleration graphs as a function of time?
An experimenter has a simple pendulum of length \( L \) and a mass–spring system with mass \( m \) and spring constant \( k \). Both are found to have the same period of oscillation \( T \) on Earth. If both systems are taken to the Moon, where the acceleration due to gravity is approximately \( \frac{1}{6} g \) of Earth, what will happen to their periods?
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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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