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 \( 4 \text{ rad/s} \)?
General form: \( a = -\omega^2 x \). Since \( \omega = 4 \), \( \omega^2 = 16 \).
Correct Answer: B
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?
\( |a_{max}| = \omega^2 A \). Given \( \omega^2 = 9 \) and \( A = 2 \).
\( |a_{max}| = 9 \times 2 = 18 \text{ m/s}^2 \).
Correct Answer: C
\( |a_{max}| = 9 \times 2 = 18 \text{ m/s}^2 \).
Correct Answer: C
Q3: Which of the following graphs correctly represents the net force \( F \) vs displacement \( x \) for an object in SHM?
SHM requires \( F = -kx \). This is a linear equation with a negative slope (\( -k \)) and a y-intercept of zero.
Correct Answer: C
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:
\( T \propto \sqrt{m} \). \( \sqrt{4} = 2 \). New period is \( 2T \).
Correct Answer: C
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?
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
Correct Answer: C
Q3: Two springs with constants \( k \) and \( 2k \) are connected in series to a mass \( m \). What is the effective period?
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
\( T = 2\pi \sqrt{\displaystyle\frac{m}{2k/3}} = 2\pi \sqrt{\displaystyle\frac{3m}{2k}} \).
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?
\( v_{max} = A\omega = 0.5 \times 10 = 5 \text{ m/s} \).
Correct Answer: B
Correct Answer: B
Q2: At \( t=0 \), a particle is at \( x = -A \). What is the phase constant \( \phi \) in the form \( x(t) = A\cos(\omega t + \phi) \)?
\( -A = A\cos(\phi) \Rightarrow \cos(\phi) = -1 \Rightarrow \phi = \pi \).
Correct Answer: C
Correct Answer: C
Q3: If the frequency is \( f \), the maximum acceleration \( a_{max} \) in terms of amplitude \( A \) is:
\( a_{max} = \omega^2 A \). Since \( \omega = 2\pi f \), \( \omega^2 = 4\pi^2 f^2 \).
\( a_{max} = 4\pi^2 f^2 A \).
Correct Answer: B
\( a_{max} = 4\pi^2 f^2 A \).
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?
\( U = \displaystyle\frac{1}{2}k(A/2)^2 = \displaystyle\frac{1}{4}(\displaystyle\frac{1}{2}kA^2) = E/4 \).
Correct Answer: B
Correct Answer: B
Q2: At what displacement \( x \) (in terms of \( A \)) is the kinetic energy exactly equal to the potential energy?
\( K = U \Rightarrow U = E/2 \).
\( \displaystyle\frac{1}{2}kx^2 = \displaystyle\frac{1}{2}(\displaystyle\frac{1}{2}kA^2) \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt{2} \).
Correct Answer: B
\( \displaystyle\frac{1}{2}kx^2 = \displaystyle\frac{1}{2}(\displaystyle\frac{1}{2}kA^2) \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt{2} \).
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:
\( E = \displaystyle\frac{1}{2}kA^2 \). If \( k \rightarrow 2k \), then \( E \rightarrow 2E \).
Correct Answer: C
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)? (\( I = \displaystyle\frac{1}{3}ML^2 \))
\( T = 2\pi \sqrt{\displaystyle\frac{ML^2/3}{Mg(L/2)}} = 2\pi \sqrt{\displaystyle\frac{2L}{3g}} \).
Correct Answer: C
Correct Answer: C
Q2: A simple pendulum has length \( L \). If decreased to \( 0.25L \), the frequency \( f \) becomes:
\( f = \displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{g}{L}} \). If \( L \rightarrow 0.25L \), \( f \rightarrow \displaystyle\frac{1}{\sqrt{0.25}}f = 2f \).
Correct Answer: B
Correct Answer: B
Q3: A uniform thin hoop of radius \( R \) is hung on a nail. What is its period of oscillation? (\( I_{rim} = 2MR^2 \))
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
\( T = 2\pi \sqrt{\displaystyle\frac{2MR^2}{MgR}} = 2\pi \sqrt{\displaystyle\frac{2R}{g}} \).
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\( T = 2\pi \sqrt{\displaystyle\frac{L}{g}} \)
