Tag: Oscillations

Unit 7: Oscillations Comprehensive Review & Advanced Problem Set Section 7.1 Defining Simple Harmonic Motion (SHM) SHM is defined by a restoring force proportional to displacement: \( F = -kx \). The differential form is \( \displaystyle\frac{d^2x}{dt^2} + \omega^2 x = 0 \). Q1: Which equation describes SHM with an angular frequency of \( 4 \text{ rad/s} \)? (A) \( \displaystyle\frac{d^2x}{dt^2} = -4x \) (B) \( \displaystyle\frac{d^2x}{dt^2} = -16x \) (C) \( \displaystyle\frac{d^2x}{dt^2} = 4x^2 \) (D) \( \displaystyle\frac{d^2x}{dt^2} = -16x^2 \) General form: \( a = -\omega^2 x \). Since \( \omega = 4 \), \( \omega^2 = 16 \). Correct Answer: B Q2: A particle undergoes motion where acceleration is \( a = -9x \). If the amplitude is \( 2 \text{ m} \), what is the magnitude of maximum acceleration? (A) \( 4.5 \text{ m/s}^2 \) (B) \( 9 \text{ m/s}^2 \) (C) \( 18 \text{ m/s}^2 \) (D) \( 36 \text{ m/s}^2 \) \( |a_{max}| = \omega^2 A \). Given \( \omega^2 = 9 \) and \( A = 2 \). \( |a_{max}| = 9 \times 2 = 18 \text{ m/s}^2 \). Correct Answer: C Q3: Which of the following graphs correctly represents the net force \( F \) vs displacement \( x \) for an object in SHM? (A) A horizontal line (B) A parabola opening upward (C) A straight line with a negative slope passing through the origin (D) A straight line with a positive slope passing through the origin SHM requires \( F = -kx \). This is a linear equation with a negative slope (\( -k \)) and a y-intercept of zero. Correct Answer: C Section 7.2 Frequency and Period of SHM The period (\( T \)) is the time for one cycle. For a mass-spring: \( T = 2\pi \sqrt{\displaystyle\frac{m}{k}} \). Q1: A mass-spring system has period \( T \). If mass is quadrupled, the new period is: (A) \( T/2 \) (B) \( T \) (C) \( 2T \) (D) \( 4T \) \( T \propto \sqrt{m} \). \( \sqrt{4} = 2 \). New period is \( 2T \). Correct Answer: C Q2: A mass-spring oscillator is moved from Earth to a planet with twice the surface gravity. How does the period change? (A) It increases (B) It decreases (C) It remains the same (D) It depends on the spring orientation The period of a mass-spring system depends only on \( m \) and \( k \) (\( T = 2\pi\sqrt{m/k} \)). It is independent of local gravity \( g \). Correct Answer: C Q3: Two springs with constants \( k \) and \( 2k \) are connected in series to a mass \( m \). What is the effective period? (A) \( 2\pi \sqrt{\displaystyle\frac{3m}{2k}} \) (B) \( 2\pi \sqrt{\displaystyle\frac{m}{3k}} \) (C) \( 2\pi \sqrt{\displaystyle\frac{2m}{3k}} \) (D) \( 2\pi \sqrt{\displaystyle\frac{m}{k}} \) In series: \( \displaystyle\frac{1}{k_{eff}} = \displaystyle\frac{1}{k} + \displaystyle\frac{1}{2k} = \displaystyle\frac{3}{2k} \Rightarrow k_{eff} = \displaystyle\frac{2k}{3} \). \( T = 2\pi \sqrt{\displaystyle\frac{m}{2k/3}} = 2\pi \sqrt{\displaystyle\frac{3m}{2k}} \). Correct Answer: A Section 7.3 Representing and Analyzing SHM Modeling SHM with \( x(t) = A\cos(\omega t + \phi) \). Q1: Position is \( x(t) = 0.5\cos(10t) \). What is the maximum speed? (A) \( 0.5 \text{ m/s} \) (B) \( 5 \text{ m/s} \) (C) \( 10 \text{ m/s} \) (D) \( 50 \text{ m/s} \) \( v_{max} = A\omega = 0.5 \times 10 = 5 \text{ m/s} \). Correct Answer: B Q2: At \( t=0 \), a particle is at \( x = -A \). What is the phase constant \( \phi \) in the form \( x(t) = A\cos(\omega t + \phi) \)? (A) \( 0 \) (B) \( \pi/2 \) (C) \( \pi \) (D) \( 3\pi/2 \) \( -A = A\cos(\phi) \Rightarrow \cos(\phi) = -1 \Rightarrow \phi = \pi \). Correct Answer: C Q3: If the frequency is \( f \), the maximum acceleration \( a_{max} \) in terms of amplitude \( A \) is: (A) \( 2\pi f A \) (B) \( 4\pi^2 f^2 A \) (C) \( 2\pi f^2 A \) (D) \( f^2 A \) \( a_{max} = \omega^2 A \). Since \( \omega = 2\pi f \), \( \omega^2 = 4\pi^2 f^2 \). \( a_{max} = 4\pi^2 f^2 A \). Correct Answer: B Section 7.4 Energy of SHM Total energy: \( E = \displaystyle\frac{1}{2}mv^2 + \displaystyle\frac{1}{2}kx^2 = \displaystyle\frac{1}{2}kA^2 \). Q1: At \( x = A/2 \), what fraction of energy is potential? (A) \( 1/2 \) (B) \( 1/4 \) (C) \( 3/4 \) (D) \( 1/\sqrt{2} \) \( U = \displaystyle\frac{1}{2}k(A/2)^2 = \displaystyle\frac{1}{4}(\displaystyle\frac{1}{2}kA^2) = E/4 \). Correct Answer: B Q2: At what displacement \( x \) (in terms of \( A \)) is the kinetic energy exactly equal to the potential energy? (A) \( A/2 \) (B) \( A/\sqrt{2} \) (C) \( A/4 \) (D) \( A\sqrt{3}/2 \) \( K = U \Rightarrow U = E/2 \). \( \displaystyle\frac{1}{2}kx^2 = \displaystyle\frac{1}{2}(\displaystyle\frac{1}{2}kA^2) \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt{2} \). Correct Answer: B Q3: A mass-spring system has total energy \( E \). If the spring constant is doubled while the amplitude remains the same, the new energy is: (A) \( E \) (B) \( \sqrt{2}E \) (C) \( 2E \) (D) \( 4E \) \( E = \displaystyle\frac{1}{2}kA^2 \). If \( k \rightarrow 2k \), then \( E \rightarrow 2E \). Correct Answer: C Section 7.5 Pendulums Simple: \( T = 2\pi \sqrt{\displaystyle\frac{L}{g}} \). Physical: \( T = 2\pi \sqrt{\displaystyle\frac{I}{mgD}} \). Q1: Period of a rod (end-pivoted)? (\( I = \displaystyle\frac{1}{3}ML^2 \)) (A) \( 2\pi \sqrt{L/g} \) (B) \( 2\pi \sqrt{L/2g} \) (C) \( 2\pi \sqrt{2L/3g} \) (D) \( 2\pi \sqrt{3L/2g} \) \( T = 2\pi \sqrt{\displaystyle\frac{ML^2/3}{Mg(L/2)}} = 2\pi \sqrt{\displaystyle\frac{2L}{3g}} \). Correct Answer: C Q2: A simple pendulum has length \( L \). If decreased to \( 0.25L \), the frequency \( f \) becomes: (A) \( f/2 \) (B) \( 2f \) (C) \( 4f \) (D) \( \sqrt{2}f \) \( f = \displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{g}{L}} \). If \( L \rightarrow 0.25L \), \( f \rightarrow \displaystyle\frac{1}{\sqrt{0.25}}f = 2f \). Correct Answer: B Q3: A uniform thin hoop of radius \( R \) is hung on a nail. What is its period of oscillation? (\( I_{rim} = 2MR^2 \)) (A) \( 2\pi \sqrt{R/g} \) (B) \( 2\pi \sqrt{2R/g} \) (C) \( 2\pi \sqrt{3R/2g} \) (D) \( 2\pi \sqrt{R/2g} \) Using the physical pendulum formula: \( I = 2MR^2 \), and \( D = R \). \( T = 2\pi \sqrt{\displaystyle\frac{2MR^2}{MgR}} = 2\pi \sqrt{\displaystyle\frac{2R}{g}} \). Correct Answer: B Oscillations Formulas Recap Angular Freq\( \omega = \sqrt{\displaystyle\frac{k}{m}} \) Period\( T = \displaystyle\frac{2\pi}{\omega} \) Position\( x(t) = A\cos(\omega t + \phi) \) Max Speed\( v_{max} = A\omega \) Total Energy\( E = \displaystyle\frac{1}{2}kA^2 \) Simple Pendulum\( T = 2\pi \sqrt{\displaystyle\frac{L}{g}} \)