Physics 1 Spring Semester Comprehensive Review Guide

Simple Harmonic Motion and Pendulums

• Period of a Simple Pendulum: The period of a pendulum, which is the time it takes for one complete oscillation, is governed by the formula $T = 2\pi\sqrt{\frac{L}{g}}$.     * Effect of Mass: If you decrease the mass of the pendulum, the period ($T$) remains unchanged. The mass of the bob does not affect the period of a simple pendulum.     * Effect of Length: If you decrease the length of the pendulum ($L$), the period ($T$) will decrease. There is a direct square-root relationship between length and period.     * Effect of Amplitude: If you decrease the amplitude of a pendulum, the period ($T$) remains unchanged, provided the angle stays relatively small (typically less than $15^{\circ}$).

• Oscillating Mass on a Spring: For a mass oscillating horizontally on a spring, the period is calculated using $T = 2\pi\sqrt{\frac{m}{k}}$.     * Problem 4 Calculation: Given a mass $m = 0.75\,kg$ and a spring constant $k = 15\,N/m$, the period is:         * $T = 2\pi\sqrt{\frac{0.75}{15}}$         * $T = 2\pi\sqrt{0.05}$         * $T \approx 1.405\,s$

• Spring Constant Calculation (Problem 28): A spring stretches based on Hooke's Law ($F = k\Delta x$). When a mass is hung vertically, gravity provides the force ($F = mg$).     * Example: Mass $m = 1.2\,kg$, Displacement $\Delta x = 5.0\,cm = 0.05\,m$.     * $k = \frac{mg}{\Delta x}$     * $k = \frac{1.2 \times 9.8}{0.05} = 235.2\,N/m$

• Pendulum Period on Earth (Problem 30): For a length $L = 1.0\,m$ and $g = 9.8\,m/s^2$:     * $T = 2\pi\sqrt{\frac{1.0}{9.8}} \approx 2.007\,s$

• Calculating Pendulum Length from Frequency (Problem 31): Given frequency $f = 0.48\,Hz$, find period $T = \frac{1}{f} = \frac{1}{0.48} \approx 2.083\,s$.     * Using $T = 2\pi\sqrt{\frac{L}{g}}$, we solve for $L = g\left(\frac{T}{2\pi}\right)^2$.     * $L = 9.8\left(\frac{2.083}{2\pi}\right)^2 \approx 1.077\,m$

• Mass on a Horizontal Spring (Problem 36): Given period $T = 0.75\,s$ and $k = 22\,N/m$, find mass $m$.     * $m = k\left(\frac{T}{2\pi}\right)^2$     * $m = 22\left(\frac{0.75}{2\pi}\right)^2 \approx 0.313\,kg$

Wave Characteristics and Standing Waves

• Standing Wave Terminology:     * Node: A point of zero amplitude on a standing wave caused by destructive interference.     * Antinode: A point of maximum amplitude on a standing wave caused by constructive interference.     * Mode Number (n): Refers to the number of harmonic loops in the standing wave.

• Wave Relationships:     * The fundamental wave equation is $v = f\lambda$, where $v$ is speed, $f$ is frequency, and $\lambda$ is wavelength.     * Speed Dependency: The speed of a wave depends on the properties of the medium, not the frequency. If a wave's frequency doubles in the same medium (the same string), the speed ($v$) remains the same ($2.5\,m/s$ in problem 7); however, the wavelength would be halved.

• Standing Wave Calculations (Problems 8–12): Based on a string length of $L = 5.0\,m$ and an illustration of a standing wave:     * Relationship: The number of loops corresponds to the mode number ($n$).     * Wavelength: $\lambda = \frac{2L}{n}$.     * Frequency: $f = \frac{nv}{2L}$.     * Fundamental Frequency: $f_1 = \frac{v}{2L}$.

• Universal Wave Example (Problem 29): For a wave with $v = 340.\,m/s$ and $f = 455\,Hz$:     * Wavelength: $\lambda = \frac{v}{f} = \frac{340.}{455} \approx 0.747\,m$     * Period: $T = \frac{1}{f} = \frac{1}{455} \approx 0.0022\,s$

Sound, Pitch, and Resonance

• Nature of Waves:     * Light vs. Sound: Light is an electromagnetic wave and does not require a medium to travel, so it can travel in a vacuum. Sound is a mechanical longitudinal wave that requires a medium (atoms/molecules) to transmit energy; therefore, it cannot travel in a vacuum.     * Transverse Wave: Particle displacement is perpendicular to the direction of wave travel (e.g., light, string waves).     * Longitudinal Wave: Particle displacement is parallel to the direction of wave travel (e.g., sound).

• Resonance: The phenomenon that occurs when the frequency of a forced vibration on an object matches that object's natural frequency, resulting in a dramatic increase in amplitude.

• Pitch and Octaves:     * Pitch: In music and acoustics, pitch is the human perception of the frequency of a sound wave.     * Octaves: Increasing a frequency by 3 octaves means doubling the frequency three times (multiplying by $2^3 = 8$).     * Example (Problem 27): 3 octaves higher than $100.\,Hz$ is $100 \times 2 \times 2 \times 2 = 800\,Hz$.

• Doppler Effect:     * When a source moves toward a stationary observer, the observer hears a higher pitch (higher frequency).     * When an observer moves away from a stationary source, the observer hears a lower pitch (lower frequency).

• Speed of Sound and Temperature: The speed of sound in air varies with temperature ($T$ in Celsius): $v = 331\,m/s + 0.6(T)$.     * Problem 21: At $30^{\circ}C$, $v = 331 + 0.6(30) = 349\,m/s$.     * Problem 35: At $20^{\circ}C$, $v = 331 + 0.6(20) = 343\,m/s$.

• Acoustic Pipes (Problems 32–33): At $25.0^{\circ}C$, $v = 331 + 0.6(25) = 346\,m/s$.     * Open Tube (Both Ends): Fundamental frequency $f_1 = \frac{v}{2L}$. To get $450\,Hz$, $L = \frac{346}{2 \times 450} \approx 0.384\,m$.     * Closed Tube (One End): Fundamental frequency $f_1 = \frac{v}{4L}$. To get $450\,Hz$, $L = \frac{346}{4 \times 450} \approx 0.192\,m$.

• Beats:     * Beats are caused by the interference of two sound waves of slightly different frequencies.     * Beat Frequency: $f_{beat} = |f_1 - f_2|$.     * Problem 20: One fork is $300\,Hz$, beat frequency is $6\,Hz$; if the second is lower, then $f_2 = 300 - 6 = 294\,Hz$.     * Problem 34: $f_1 = 400.\,Hz$, $f_2 = 405\,Hz$.         * $f_{beat} = 5\,Hz$.         * Period of beat $T_{beat} = \frac{1}{f_{beat}} = \frac{1}{5} = 0.2\,s$.

• Perceived Frequency Moving Away (Problem 35): $f_{obs} = f_s \left( \frac{v}{v + v_s} \right)$ for a source moving away.     * $f_{obs} = 350 \left( \frac{343}{343 + 45} \right) \approx 309.4\,Hz$.

Electromagnetic Spectrum and Color

• EM Spectrum Regions:     * Longest Wavelength: Radio waves.     * Highest Frequency: Gamma rays.

• Visible Light:     * Highest Energy per Photon: Violet light (highest frequency corresponds to highest energy, $E = hf$).     * Primary Colors of Light: Red, Green, Blue.     * Complementary Colors of Light:         * Red + Cyan = White (Cyan is complement of Red).         * Green + Magenta = White (Magenta is complement of Green).         * Blue + Yellow = White (Yellow is complement of Blue).     * Primary Pigments: Cyan, Magenta, Yellow.

Geometric Optics: Mirrors and Lenses

• Concave Mirror Scenarios:     * Object between focal point (f) and center of curvature (C): Image is real, inverted, magnified, and appears beyond C.     * Object on focal point (f): No image is formed (rays are parallel).     * Object on center of curvature (C): Image is real, inverted, same size, and appears at C.

• Converging Lens Scenarios:     * Between f and 2f (Center of Curvature): Image is real, inverted, magnified, and appears beyond 2f on the other side.     * On focal point (f): No image formed.     * On 2f: Image is real, inverted, same size, and appears at 2f on the other side.

• Mathematical Calculations for Mirrors (Problem 38): Object height $h_o = 3.5\,cm$, Object distance $d_o = 4.5\,cm$, Radius $R = 10.0\,cm$.     * Focal length: $f = \frac{R}{2} = 5.0\,cm$.     * Image distance ($d_i$): $\frac{1}{5.0} = \frac{1}{4.5} + \frac{1}{d_i} \Rightarrow \frac{1}{d_i} = 0.2 - 0.222 = -0.0222 \Rightarrow d_i = -45\,cm$. (Virtual, upright, magnified image).     * If object is at $d_o = 25\,cm$: $\frac{1}{5.0} = \frac{1}{25} + \frac{1}{d_i} \Rightarrow \frac{1}{d_i} = 0.2 - 0.04 = 0.16 \Rightarrow d_i = 6.25\,cm$. (Real, inverted, reduced image).

• Refraction and Critical Angle (Problem 42):     * Light travels from Pepsi ($n_1 = 1.40$) to air ($n_2 = 1.00$) at $\theta_1 = 23^{\circ}$.     * Snell's Law: $1.40 \times \sin(23^{\circ}) = 1.00 \times \sin(\theta_2)$.     * $\sin(\theta_2) = 1.40 \times 0.3907 = 0.547 \Rightarrow \theta_2 \approx 33.16^{\circ}$.     * Critical Angle: $\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) = \sin^{-1}\left(\frac{1.00}{1.40}\right) \approx 45.6^{\circ}$.

• Lens Calculation (Problem 43): Converging lens, $f = 15\,cm$, $d_o = 22\,cm$.     * $\frac{1}{15} = \frac{1}{22} + \frac{1}{d_i} \Rightarrow \frac{1}{d_i} = 0.0667 - 0.0454 = 0.0213 \Rightarrow d_i \approx 46.9\,cm$.

Universal Gravitation, Electrostatics, and Circuits

• Coulomb's Law (Problem 44):     * $q_1 = +5.8 \times 10^{-6}\,C$, $q_2 = -1.6 \times 10^{-6}\,C$, $r = 1.46\,mm = 1.46 \times 10^{-3}\,m$.     * $F = k \frac{|q_1q_2|}{r^2} = (8.99 \times 10^9) \frac{(5.8 \times 10^{-6})(1.6 \times 10^{-6})}{(1.46 \times 10^{-3})^2}$     * $F \approx 3.91 \times 10^4\,N$. Force is attractive because charges are opposite.

• Satellite Motion (Problem 45):     * Altitude $h = 150\,km = 0.15 \times 10^6\,m$ . Total radius $r = R_E + h = 6.37 \times 10^6 + 0.15 \times 10^6 = 6.52 \times 10^6\,m$.     * Linear Speed: $v = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.52 \times 10^6}} \approx 7822\,m/s$.     * Period: $T = \frac{2\pi r}{v} = \frac{2\pi(6.52 \times 10^6)}{7822} \approx 5237\,s$.

• Circuits (Problems 46–49):     * Series Resistance: $R_{eq} = R_1 + R_2 + …$     * Parallel Resistance: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + …$     * Ohm's Law: $V = IR$.     * Voltage and Current Rules:         * In series: Current is the same everywhere; voltages add up to the total source voltage.         * In parallel: Voltage is the same across all branches; currents add up to the total current.