Welcome to Lecture 111 (Master Index Entry: L105) of our comprehensive JEE Physics Master Series! Today, we inaugurate Unit 10: Oscillations and Waves with an essential exploration of Oscillations and Periodic Motion. Building from fundamental kinematics into vibrational mechanics, we analyze the rigorous criteria distinguishing general periodic motion from true oscillatory motion ($F = -k x^n$), define fundamental kinematic variables (time period $T$, frequency $f$, and angular frequency $\omega$), examine Fourier decomposition, derive harmonic displacement equations as functions of time ($x(t) = A \sin(\omega t + \phi)$), evaluate the phasor method for superposition, investigate commensurate vs. incommensurate frequencies, and master high-yield JEE Advanced problem archetypes.
Detailed Concept Card: Oscillations & Kinematics of Periodic Motion
1. Periodic Motion vs. Oscillatory Motion
Understanding the precise mechanical hierarchy between periodic and oscillatory phenomena is crucial for competitive exams:
- Periodic Motion: Any motion that identically repeats itself at regular, equal intervals of time ($T$). Mathematically, the position coordinate satisfies: $$f(t + T) = f(t)$$ Examples: Uniform circular motion of planets around the Sun, the rotation of the Earth on its axis, hands of an analog clock. Periodic motion does NOT require a restoring force or a back-and-forth trajectory!
- Oscillatory (or Vibratory) Motion: A special class of periodic motion where a body moves to-and-fro (back-and-forth) about a fixed mean position (stable equilibrium). Condition for Stable Oscillation: The system must experience a restoring force (or torque) that always acts in opposition to displacement from equilibrium: $$F(x) = -k x^n$$ – When $n = 1, 3, 5\dots$ (any odd positive integer), $F(x)$ is directed opposite to $x$ on both sides of equilibrium, producing stable oscillations. – When $n = 1$ ($F = -kx$), the acceleration is directly proportional to displacement ($a \propto -x$), defining Simple Harmonic Motion (SHM). – When $n$ is even ($n = 2, 4\dots$), $F(x)$ acts in the same direction regardless of the sign of $x$, making stable oscillatory motion impossible!
- Universal Rule: All oscillatory motions are periodic, but not all periodic motions are oscillatory!
2. Kinematic Variables of Oscillatory Motion
- Time Period ($T$): The smallest time interval after which the motion repeats its state (position, velocity, and acceleration): $$T = \frac{1}{f} = \frac{2\pi}{\omega}$$
- Frequency ($f$ or $\nu$): The number of complete cycles executed per second: $$f = \frac{1}{T} \quad (\text{SI Unit: Hertz, Hz} = \text{s}^{-1})$$
- Angular Frequency ($\omega$): The rate of change of phase angle: $$\omega = 2\pi f = \frac{2\pi}{T} \quad (\text{SI Unit: rad/s})$$
- Displacement ($x(t)$ or $y(t)$): The instantaneous directed distance of the oscillating particle from its central mean equilibrium position.
- Amplitude ($A$): The maximum magnitude of displacement from the mean position ($|x|_{\text{max}} = A$).
3. Mathematical Representation: Harmonic vs. Non-Harmonic Functions
By Fourier’s Theorem, any bounded, single-valued periodic function of period $T$ can be decomposed into an infinite series of sinusoidal functions:
$$x(t) = A_0 + \sum_{n=1}^\infty \left[A_n \cos(n \omega t) + B_n \sin(n \omega t)\right]$$
- Simple Harmonic Functions: A single sinusoidal term represents pure SHM: $$x(t) = A \sin(\omega t + \phi) \quad \text{or} \quad x(t) = A \cos(\omega t – \delta)$$
- Linear Superposition of Same Frequency: $$x(t) = a \sin(\omega t) + b \cos(\omega t) = A \sin(\omega t + \phi)$$ where the resultant amplitude is $A = \sqrt{a^2 + b^2}$, and the initial phase is $\tan\phi = b/a$. This is strictly SHM!
- Powers of Harmonic Functions: Functions like $x(t) = \sin^2(\omega t)$ are simple harmonic about a shifted mean position: $$x(t) = \frac{1 – \cos(2\omega t)}{2} = \underbrace{\frac{1}{2}}_{\text{Shifted Mean}} – \underbrace{\frac{1}{2} \cos(2\omega t)}_{\text{SHM of amplitude } 1/2 \text{ and period } \pi/\omega}$$
- Multiple Harmonic Combinations: $x(t) = a_1 \sin(\omega_1 t) + a_2 \sin(\omega_2 t)$ is: – Periodic: If and only if the frequency ratio $\frac{\omega_1}{\omega_2}$ is a rational number ($p/q$). The time period is $T = \text{LCM}(T_1, T_2)$. – Non-Periodic (Quasi-Periodic): If $\frac{\omega_1}{\omega_2}$ is an irrational number, the motion never repeats identically!
4. General SHM Displacement Equation & Phase Interpretation
The general displacement equation is formulated as:
$$\mathbf{x(t) = A \sin(\omega t + \phi_0)}$$
- Phase $\theta(t) = \omega t + \phi_0$: Completely describes both the instantaneous position and direction of motion of the particle.
- Initial Phase (Epoch, $\phi_0$): The phase value at $t = 0$:
- If particle starts at mean position moving towards $+A$ at $t = 0$: $x(0) = 0, v(0) > 0 \implies \mathbf{\phi_0 = 0} \implies x(t) = A \sin(\omega t)$.
- If particle starts at positive extreme at $t = 0$: $x(0) = +A \implies \mathbf{\phi_0 = \pi/2} \implies x(t) = A \cos(\omega t)$.
- If particle starts at negative extreme at $t = 0$: $x(0) = -A \implies \mathbf{\phi_0 = 3\pi/2} \implies x(t) = -A \cos(\omega t)$.
Four Graded Solved Examples
Example 1 (JEE Main Level: Identifying Periodic & Simple Harmonic Motion)
Problem Statement: Analyze each of the following mathematical functions of time. Classify each as (i) periodic and SHM, (ii) periodic but not SHM, or (iii) non-periodic. For periodic functions, evaluate the fundamental time period $T$ in terms of $\omega$:
1. $x_1(t) = \sin(\omega t) + \cos(\omega t)$
2. $x_2(t) = \sin^2(\omega t)$
3. $x_3(t) = \sin(\omega t) + \sin(2\omega t) + \sin(4\omega t)$
4. $x_4(t) = e^{-\omega t}$
5. $x_5(t) = \log(\omega t)$
Solution:
Analysis of 1: $x_1(t) = \sin(\omega t) + \cos(\omega t)$
Multiply and divide by $\sqrt{2}$:
$$x_1(t) = \sqrt{2} \left[\frac{1}{\sqrt{2}} \sin(\omega t) + \frac{1}{\sqrt{2}} \cos(\omega t)\right] = \sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right)$$
This is a single sinusoidal function of angular frequency $\omega$.
Classification: Periodic and Simple Harmonic Motion (SHM).
Amplitude $A = \sqrt{2}$, Time period $\mathbf{T = \frac{2\pi}{\omega}}$.
Analysis of 2: $x_2(t) = \sin^2(\omega t)$
Using the trigonometric identity $\sin^2(\theta) = \frac{1 – \cos(2\theta)}{2}$:
$$x_2(t) = \frac{1}{2} – \frac{1}{2} \cos(2\omega t)$$
This represents simple harmonic oscillation of amplitude $1/2$ about a shifted mean equilibrium position $x = 1/2$ with angular frequency $\omega’ = 2\omega$.
Classification: Periodic and SHM (about shifted origin $x = 1/2$).
Time period $\mathbf{T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}}$.
Analysis of 3: $x_3(t) = \sin(\omega t) + \sin(2\omega t) + \sin(4\omega t)$
The individual periods are $T_1 = \frac{2\pi}{\omega}$, $T_2 = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$, and $T_3 = \frac{2\pi}{4\omega} = \frac{\pi}{2\omega}$.
Since all frequencies are commensurate integer multiples of $\omega$, the function is periodic with fundamental period equal to the period of the lowest harmonic ($T = 2\pi/\omega$). However, because it contains higher harmonics ($2\omega, 4\omega$), the restoring force is not linearly proportional to displacement ($d^2 x/dt^2 \neq -\omega^2 x$).
Classification: Periodic but NOT Simple Harmonic.
Time period $\mathbf{T = \frac{2\pi}{\omega}}$.
Analysis of 4: $x_4(t) = e^{-\omega t}$
As $t$ increases, $e^{-\omega t}$ decreases monotonically towards $0$. It never repeats any value.
Classification: Non-periodic.
Analysis of 5: $x_5(t) = \log(\omega t)$
The logarithmic function increases monotonically without bound as $t \to \infty$. It never repeats.
Classification: Non-periodic.
Example 2 (Mathematical Formulation: Phasor Superposition of Collinear Displacements)
Problem Statement: A particle of mass $m = 0.20\text{ kg}$ is simultaneously subjected to two collinear simple harmonic displacements of identical angular frequency $\omega = 10\text{ rad/s}$ along the $x$-axis:
$$x_1(t) = 3.0 \sin(10 t) \quad \text{and} \quad x_2(t) = 4.0 \sin\left(10 t + \frac{\pi}{3}\right)$$
where displacements are in centimeters and $t$ in seconds.
1. Find the amplitude $A_R$ of the resultant oscillation.
2. Determine the initial phase constant $\phi_R$ of the resultant motion expressed as $x_R(t) = A_R \sin(10 t + \phi_R)$.
3. Calculate the maximum speed $v_{\text{max}}$ and maximum acceleration $a_{\text{max}}$ of the particle.
Solution:
Step 1: Resultant Amplitude via Phasor Method.
We represent each displacement by a vector (phasor) in the complex plane:
Phasor 1: Amplitude $A_1 = 3.0\text{ cm}$, phase angle $\theta_1 = 0$.
Phasor 2: Amplitude $A_2 = 4.0\text{ cm}$, phase angle $\theta_2 = \pi/3 = 60^\circ$.
The phase difference between the two motions is $\Delta \phi = 60^\circ – 0^\circ = 60^\circ$.
$$A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)}$$
$$A_R = \sqrt{3.0^2 + 4.0^2 + 2(3.0)(4.0)\cos(60^\circ)} = \sqrt{9.0 + 16.0 + 24.0(0.50)} = \sqrt{25.0 + 12.0} = \mathbf{\sqrt{37}\text{ cm} \approx 6.08\text{ cm}}$$
Step 2: Resultant Phase Constant ($\phi_R$).
$$\tan\phi_R = \frac{A_2 \sin(\Delta \phi)}{A_1 + A_2 \cos(\Delta \phi)} = \frac{4.0 \sin(60^\circ)}{3.0 + 4.0 \cos(60^\circ)} = \frac{4.0 (\sqrt{3}/2)}{3.0 + 4.0(0.50)} = \frac{2\sqrt{3}}{3.0 + 2.0} = \frac{2\sqrt{3}}{5} \approx 0.6928$$
$$\phi_R = \arctan(0.6928) \approx 34.7^\circ \approx \mathbf{0.606\text{ radians}}$$
The resultant displacement equation is:
$$x_R(t) = 6.08 \sin(10 t + 0.606)\text{ cm}$$
Step 3: Maximum Velocity & Acceleration.
In SI units ($A_R = 0.0608\text{ m}, \omega = 10\text{ rad/s}$):
$$v_{\text{max}} = \omega A_R = (10\text{ rad/s})(0.0608\text{ m}) = \mathbf{0.608\text{ m/s}}$$
$$a_{\text{max}} = \omega^2 A_R = (10\text{ rad/s})^2 (0.0608\text{ m}) = \mathbf{6.08\text{ m/s}^2}$$
Example 3 (JEE Advanced Level: Time-Splitting Across Displacement Intervals)
Problem Statement: A particle executes simple harmonic motion of amplitude $A$ and time period $T$ along the $x$-axis, starting from the mean position at $t = 0$ with positive velocity ($x(t) = A \sin(\omega t)$).
1. Find the minimum time $t_1$ required to move from the mean position ($x = 0$) to half the amplitude ($x = A/2$).
2. Find the minimum time $t_2$ required to move from half the amplitude ($x = A/2$) to the positive extreme ($x = A$).
3. Explain physically why $t_2 = 2 t_1$ even though both intervals cover the exact same distance of $A/2$.
4. Determine the time required to travel from $x = 0$ to $x = A / \sqrt{2}$.
Solution:
Step 1: Time $t_1$ to reach $x = A/2$.
$$x(t_1) = A \sin(\omega t_1) = \frac{A}{2} \implies \sin(\omega t_1) = \frac{1}{2}$$
The principal value for initial motion is:
$$\omega t_1 = \frac{\pi}{6} \implies \left(\frac{2\pi}{T}\right) t_1 = \frac{\pi}{6} \implies \mathbf{t_1 = \frac{T}{12}}$$
Step 2: Time $t_2$ to travel from $x = A/2$ to $x = A$.
The time $t_{\text{ext}}$ to reach the positive extreme position ($x = A$) from $x = 0$ is:
$$\sin(\omega t_{\text{ext}}) = 1 \implies \omega t_{\text{ext}} = \frac{\pi}{2} \implies t_{\text{ext}} = \frac{T}{4}$$
The time taken to traverse from $A/2$ to $A$ is:
$$t_2 = t_{\text{ext}} – t_1 = \frac{T}{4} – \frac{T}{12} = \frac{3T – T}{12} = \frac{2T}{12} = \mathbf{\frac{T}{6}}$$
Step 3: Physical Explanation of $t_2 / t_1 = 2$.
Notice that:
$$\frac{t_2}{t_1} = \frac{T/6}{T/12} = \mathbf{2}$$
In simple harmonic motion, the speed of the particle is maximum at the mean position ($v = \omega A$) and drops to zero at the extreme position ($v = 0$ at $x = A$). Over the first half ($0 \to A/2$), the particle moves at high speeds ($v \in [\frac{\sqrt{3}}{2} \omega A, \, \omega A]$). Over the second half ($A/2 \to A$), the particle decelerates towards rest ($v \in [0, \, \frac{\sqrt{3}}{2} \omega A]$). Because the average velocity in the second half is half of that in the first half, it takes exactly twice as long to cover the same distance!
Step 4: Time to reach $x = A / \sqrt{2}$.
$$\sin(\omega t) = \frac{1}{\sqrt{2}} \implies \omega t = \frac{\pi}{4} \implies \left(\frac{2\pi}{T}\right) t = \frac{\pi}{4} \implies \mathbf{t = \frac{T}{8}}$$
Example 4 (JEE Advanced Level: Commensurate vs. Incommensurate Frequency Superposition)
Problem Statement: Consider two composite particle displacement functions:
$$y_1(t) = \sin(10\pi t) + \sin(15\pi t)$$
$$y_2(t) = \sin(10\pi t) + \sin(10\sqrt{2}\pi t)$$
1. Determine whether $y_1(t)$ and $y_2(t)$ represent periodic or non-periodic motions.
2. For the periodic function, calculate the fundamental angular frequency $\omega_0$ and the fundamental time period $T$.
3. Express $y_1(t)$ in envelope-modulated (beat) form and find the beat period.
Solution:
Step 1: Commensurability Test.
– For $y_1(t)$: The angular frequencies are $\omega_1 = 10\pi\text{ rad/s}$ and $\omega_2 = 15\pi\text{ rad/s}$.
$$\frac{\omega_1}{\omega_2} = \frac{10\pi}{15\pi} = \frac{2}{3}$$
Since $2/3$ is a rational number, there exist integers $n_1 = 3$ and $n_2 = 2$ such that $n_1 T_1 = n_2 T_2$. Therefore, $y_1(t)$ is strictly periodic.
– For $y_2(t)$: The angular frequencies are $\omega_1 = 10\pi\text{ rad/s}$ and $\omega_2 = 10\sqrt{2}\pi\text{ rad/s}$.
$$\frac{\omega_1}{\omega_2} = \frac{10\pi}{10\sqrt{2}\pi} = \frac{1}{\sqrt{2}}$$
Since $1/\sqrt{2}$ is an irrational number, no pair of integers can satisfy $n_1 T_1 = n_2 T_2$. The function never repeats itself identically over any finite duration. Therefore, $y_2(t)$ is non-periodic (quasi-periodic).
Step 2: Fundamental Period of $y_1(t)$.
The individual periods are:
$$T_1 = \frac{2\pi}{10\pi} = \frac{1}{5}\text{ s} = 0.20\text{ s}, \qquad T_2 = \frac{2\pi}{15\pi} = \frac{2}{15}\text{ s}$$
The fundamental period $T$ is the least common multiple (LCM) of $T_1$ and $T_2$:
$$T = \text{LCM}\left(\frac{1}{5}, \frac{2}{15}\right) = \frac{\text{LCM}(1, 2)}{\text{GCD}(5, 15)} = \frac{2}{5}\text{ s} = \mathbf{0.40\text{ s}}$$
The fundamental angular frequency is:
$$\omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2/5} = \mathbf{5\pi\text{ rad/s}}$$
Step 3: Beat Product Formulation.
Using the sum-to-product trigonometric identity $\sin A + \sin B = 2 \cos\left(\frac{A – B}{2}\right) \sin\left(\frac{A + B}{2}\right)$:
$$y_1(t) = 2 \cos\left(\frac{15\pi t – 10\pi t}{2}\right) \sin\left(\frac{15\pi t + 10\pi t}{2}\right) = \mathbf{2 \cos(2.5\pi t) \sin(12.5\pi t)}$$
The carrier frequency is $12.5\pi\text{ rad/s}$, and the amplitude envelope modulates at $\omega_{\text{mod}} = 2.5\pi\text{ rad/s}$. The beat frequency is $f_b = |f_2 – f_1| = \frac{15\pi – 10\pi}{2\pi} = 2.5\text{ Hz}$, corresponding to a beat period $T_b = \frac{1}{2.5} = \mathbf{0.40\text{ s}}$.
Practice Worksheet: 10 High-Yield JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. A particle executes simple harmonic motion with amplitude $A$ and time period $T$. The minimum time taken by the particle to travel from its mean equilibrium position ($x = 0$) to half of its maximum displacement ($x = A/2$) is:
(A) $\frac{T}{12}$
(B) $\frac{T}{8}$
(C) $\frac{T}{6}$
(D) $\frac{T}{4}$
Q2. The displacement of a particle as a function of time is given by $x(t) = A \sin^2(\omega t)$. The motion is:
(A) Non-periodic
(B) Periodic and simple harmonic with time period $\frac{\pi}{\omega}$
(C) Periodic and simple harmonic with time period $\frac{2\pi}{\omega}$
(D) Periodic but not simple harmonic
Q3. A particle executes SHM described by $x(t) = A \sin(\omega t + \phi)$. The maximum speed attained by the particle during its oscillation is:
(A) $\frac{\omega}{A}$
(B) $\frac{A}{\omega}$
(C) $\omega A$
(D) $\omega^2 A$
Q4. A body executes simple harmonic motion of amplitude $A$. At what displacement from the mean position is its instantaneous speed equal to half of its maximum speed ($v = v_{\text{max}} / 2$)?
(A) $\pm \frac{A}{2}$
(B) $\pm \frac{A}{\sqrt{2}}$
(C) $\pm \frac{\sqrt{3}}{2} A$
(D) $\pm \frac{2}{3} A$
Q5. For a particle executing simple harmonic motion, the phase difference between instantaneous displacement $x(t)$ and instantaneous acceleration $a(t)$ is:
(A) $0\text{ rad}$
(B) $\frac{\pi}{4}\text{ rad}$
(C) $\frac{\pi}{2}\text{ rad}$
(D) $\pi\text{ rad}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding oscillatory and periodic motions is/are correct?
(A) Every oscillatory motion is necessarily periodic, but every periodic motion is not necessarily oscillatory.
(B) For a restoring force of the form $F = -k x^n$, stable oscillatory motion occurs if and only if $n$ is an odd positive integer ($n = 1, 3, 5\dots$).
(C) The function $x(t) = \sin(\omega t) + \cos(2\omega t)$ is periodic with period $T = \frac{2\pi}{\omega}$, but does not represent simple harmonic motion.
(D) The superposition of two simple harmonic motions whose frequencies have an irrational ratio produces a non-periodic motion.
Q7. A particle executes simple harmonic motion along the $x$-axis according to $x(t) = A \cos(\omega t)$. Which of the following statements is/are correct?
(A) At time $t = 0$, the particle is located at its positive extreme position ($x = +A$).
(B) The magnitude of acceleration is maximum at the extreme positions ($x = \pm A$) and zero at the mean position ($x = 0$).
(C) The speed of the particle is maximum at the mean position ($x = 0$) and zero at the extreme positions ($x = \pm A$).
(D) The phase difference between velocity and acceleration is $\frac{\pi}{2}\text{ rad}$.
Q8. A particle is subjected to two mutually perpendicular harmonic displacements: $x(t) = A \sin(\omega t)$ and $y(t) = B \sin(\omega t + \delta)$. Which of the following statements is/are correct regarding the resulting trajectory in the $xy$-plane?
(A) If $\delta = 0$, the trajectory is a straight line through the origin with positive slope ($y = \frac{B}{A} x$).
(B) If $\delta = \pi$, the trajectory is a straight line through the origin with negative slope ($y = -\frac{B}{A} x$).
(C) If $\delta = \frac{\pi}{2}$ and $A = B$, the trajectory is a circle of radius $A$ centered at the origin ($x^2 + y^2 = A^2$).
(D) If $\delta = \frac{\pi}{2}$ and $A \neq B$, the trajectory is an ellipse whose principal axes coincide with the coordinate axes ($\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$).
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A particle executes simple harmonic motion with amplitude $A = 10\text{ cm}$ and time period $T = 6.0\text{ s}$. The minimum time taken by the particle to travel from $x = -5.0\text{ cm}$ to $x = +5.0\text{ cm}$ across the mean position in seconds is $t$. Determine the integer value of $t$.
Q10. A particle is simultaneously acted upon by two collinear harmonic motions given by $x_1(t) = 5.0 \sin(\omega t)$ and $x_2(t) = 12.0 \cos(\omega t)$ in the same units. The amplitude of the resultant displacement is $A_R$. Determine the integer value of $A_R$.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $\frac{T}{12}$ | SHM Travel Time from $x = 0$ to $x = A/2$ |
| Q2 | (B) Periodic & SHM with period $\frac{\pi}{\omega}$ | Trigonometric Decomposition of $\sin^2(\omega t)$ |
| Q3 | (C) $\omega A$ | Maximum Velocity in Simple Harmonic Motion |
| Q4 | (C) $\pm \frac{\sqrt{3}}{2} A$ | Displacement for Half-Maximum Speed |
| Q5 | (D) $\pi\text{ rad}$ | Phase Difference Between Displacement & Acceleration |
| Q6 | (A), (B), (C), (D) | Foundations of Periodic & Oscillatory Motions |
| Q7 | (A), (B), (C), (D) | Kinematic Velocity & Acceleration Profiles in SHM |
| Q8 | (A), (B), (C), (D) | Lissajous Trajectories for Perpendicular Superpositions |
| Q9 | 1 | Symmetric Interval Travel Time ($t = T/6 = 1\text{ s}$) |
| Q10 | 13 | Orthogonal Superposition Amplitude ($\sqrt{5^2 + 12^2} = 13$) |
Complete Step-by-Step Solutions
Solution 1:
Let the displacement equation starting from the mean position be $x(t) = A \sin(\omega t)$.
When $x = A/2$:
$$A \sin(\omega t) = \frac{A}{2} \implies \sin(\omega t) = \frac{1}{2} \implies \omega t = \frac{\pi}{6}$$
Since $\omega = \frac{2\pi}{T}$:
$$\left(\frac{2\pi}{T}\right) t = \frac{\pi}{6} \implies t = \frac{T}{12}$$
Correct Option: (A)
Solution 2:
Using the identity $\sin^2(\omega t) = \frac{1 – \cos(2\omega t)}{2}$:
$$x(t) = \frac{A}{2} – \frac{A}{2} \cos(2\omega t)$$
This represents simple harmonic motion of amplitude $A/2$ oscillating about the mean position $x = A/2$ with angular frequency $\omega’ = 2\omega$.
The time period is:
$$T = \frac{2\pi}{\omega’} = \frac{2\pi}{2\omega} = \mathbf{\frac{\pi}{\omega}}$$
Correct Option: (B)
Solution 3:
Velocity is the time derivative of displacement:
$$v(t) = \frac{dx}{dt} = \omega A \cos(\omega t + \phi)$$
The maximum speed occurs when $|\cos(\omega t + \phi)| = 1$:
$$v_{\text{max}} = \mathbf{\omega A}$$
Correct Option: (C)
Solution 4:
The velocity of a particle executing SHM as a function of displacement $x$ is:
$$v = \omega \sqrt{A^2 – x^2}$$
Given $v = \frac{1}{2} v_{\text{max}} = \frac{1}{2} \omega A$:
$$\omega \sqrt{A^2 – x^2} = \frac{1}{2} \omega A \implies A^2 – x^2 = \frac{A^2}{4}$$
$$x^2 = A^2 – \frac{A^2}{4} = \frac{3}{4} A^2 \implies \mathbf{x = \pm \frac{\sqrt{3}}{2} A}$$
Correct Option: (C)
Solution 5:
Let displacement be $x(t) = A \sin(\omega t)$.
Velocity is $v(t) = \frac{dx}{dt} = \omega A \cos(\omega t) = \omega A \sin\left(\omega t + \frac{\pi}{2}\right)$.
Acceleration is $a(t) = \frac{dv}{dt} = -\omega^2 A \sin(\omega t) = \omega^2 A \sin(\omega t + \pi)$.
The phase difference between $a(t)$ and $x(t)$ is $(\omega t + \pi) – (\omega t) = \mathbf{\pi\text{ radians}}$ ($180^\circ$). Acceleration is in antiphase with displacement.
Correct Option: (D)
Solution 6:
(A) Correct: Oscillation requires periodic to-and-fro motion, whereas periodic motion includes non-oscillatory paths like circular orbits.
(B) Correct: For restoring behavior $F \propto -x$, $n$ must be odd so that $F < 0$ when $x > 0$ and $F > 0$ when $x < 0$.
(C) Correct: It is a superposition of $\omega$ and $2\omega$, periodic with fundamental frequency $\omega$ ($T = 2\pi/\omega$), but the acceleration is not simply $-\omega^2 x$.
(D) Correct: If the frequency ratio is irrational, no common period exists, producing a non-periodic motion.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: $x(0) = A \cos(0) = +A$.
(B) Correct: $|a| = \omega^2 |x|$. At $x = \pm A$, $|a| = \omega^2 A$ (maximum); at $x = 0$, $a = 0$.
(C) Correct: $|v| = \omega \sqrt{A^2 – x^2}$. At $x = 0$, $v = \omega A$ (maximum); at $x = \pm A$, $v = 0$.
(D) Correct: Velocity leads displacement by $\pi/2$, and acceleration leads velocity by $\pi/2$, so the phase difference between velocity and acceleration is $\pi/2\text{ rad}$.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: For $\delta = 0$, $\frac{y}{B} = \frac{x}{A} \implies y = \left(\frac{B}{A}\right) x$ (straight line in 1st and 3rd quadrants).
(B) Correct: For $\delta = \pi$, $\frac{y}{B} = -\frac{x}{A} \implies y = -\left(\frac{B}{A}\right) x$ (straight line in 2nd and 4th quadrants).
(C) Correct: For $\delta = \pi/2$ and $A = B$, $x = A \sin(\omega t), y = A \cos(\omega t) \implies x^2 + y^2 = A^2$ (circle).
(D) Correct: For $\delta = \pi/2$ and $A \neq B$, $x = A \sin(\omega t), y = B \cos(\omega t) \implies \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ (ellipse).
Correct Options: (A), (B), (C), (D)
Solution 9:
The time required to travel from $x = -A/2$ to $x = 0$ is $T/12$.
The time required to travel from $x = 0$ to $x = +A/2$ is $T/12$.
The minimum total time required to travel from $x = -5.0\text{ cm}$ to $x = +5.0\text{ cm}$ is:
$$t = \frac{T}{12} + \frac{T}{12} = \frac{T}{6}$$
Given $T = 6.0\text{ s}$:
$$t = \frac{6.0\text{ s}}{6} = \mathbf{1\text{ s}}$$
Answer: 1
Solution 10:
Rewrite $x_2(t)$ in sine form:
$$x_2(t) = 12.0 \cos(\omega t) = 12.0 \sin\left(\omega t + \frac{\pi}{2}\right)$$
The phase difference between $x_1(t)$ and $x_2(t)$ is $\Delta \phi = \frac{\pi}{2} = 90^\circ$.
The resultant amplitude is:
$$A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(90^\circ)} = \sqrt{A_1^2 + A_2^2}$$
$$A_R = \sqrt{5.0^2 + 12.0^2} = \sqrt{25.0 + 144.0} = \sqrt{169.0} = \mathbf{13}$$
Answer: 13