Phase in Simple Harmonic Motion (SHM) & Graphical Representation: x-t, v-t, a-t Curves, Phasors & JEE Problems | Class 11 Physics

Welcome to Lecture 113 (Master Index Entry: L107) of our comprehensive JEE Physics Master Series! Continuing in Unit 10: Oscillations and Waves, we explore the angular and geometric heart of oscillatory motion: Phase in Simple Harmonic Motion and its Graphical Representations. Phase serves as the master kinematic coordinate of SHM, encoding both the instantaneous position and the instantaneous direction of motion. We analyze the physical meaning of phase ($\omega t + \phi_0$) and epoch ($\phi_0$), examine the reference circle (phasor) method of uniform circular motion projection, construct and compare the $x-t, v-t,$ and $a-t$ harmonic curves, explore phase-space trajectories in the $v-x$ and $a-x$ planes, investigate Lissajous perpendicular superpositions, and master high-yield JEE Advanced problem patterns.

Detailed Concept Card: Phase, Reference Circle & Graphical Kinematics

1. The Concept of Phase & Epoch

In the general displacement equation for simple harmonic motion:

$$\mathbf{x(t) = A \sin(\omega t + \phi_0)}$$

• Instantaneous Phase ($\theta(t) = \omega t + \phi_0$): The entire argument of the periodic function. Measured in radians, it completely specifies the physical state of the oscillator at any instant $t$, including:
  1. The position (displacement $x$) of the particle relative to the mean position.
  2. The direction of velocity (whether moving in the $+x$ or $-x$ direction).
• Initial Phase (Epoch / Phase Constant, $\phi_0$): The value of the phase angle at the initial instant $t = 0$. It depends purely on the choice of the time origin and initial boundary conditions:
  • Mean Position Moving in $+x$ Direction: $x(0) = 0, v(0) > 0 \implies \mathbf{\phi_0 = 0} \implies x(t) = A \sin(\omega t)$.
  • Positive Extreme Position: $x(0) = +A, v(0) = 0 \implies \mathbf{\phi_0 = \frac{\pi}{2}} \implies x(t) = A \cos(\omega t)$.
  • Mean Position Moving in $-x$ Direction: $x(0) = 0, v(0) < 0 \implies \mathbf{\phi_0 = \pi} \implies x(t) = -A \sin(\omega t)$.
  • Negative Extreme Position: $x(0) = -A, v(0) = 0 \implies \mathbf{\phi_0 = \frac{3\pi}{2}} \implies x(t) = -A \cos(\omega t)$.
• Phase Difference ($\Delta \phi$): For two SHMs of identical angular frequency $\omega$ given by $x_1 = A_1 \sin(\omega t + \phi_1)$ and $x_2 = A_2 \sin(\omega t + \phi_2)$, the phase difference is: $$\Delta \phi = (\omega t + \phi_2) – (\omega t + \phi_1) = \phi_2 – \phi_1 = \text{constant}$$ Because $\Delta \phi$ is time-independent, the two oscillators maintain a fixed phase relationship throughout their motion.

2. The Reference Circle (Phasor) Method

Simple Harmonic Motion can be geometrically constructed as the one-dimensional orthogonal projection of uniform circular motion (UCM) onto a diameter.

Parameter / Feature	Reference Circle (UCM)	Projected SHM (Diameter)
Radius	Radius of reference circle $R$	Amplitude $A = R$
Angular Speed	Uniform orbital angular speed $\omega$	Oscillation angular frequency $\omega = \sqrt{k/m}$
Position Vector	Phasor $\vec{r}(t)$ making angle $\theta = \omega t + \phi_0$ with reference axis	$x(t) = A \sin(\omega t + \phi_0)$ (or cosine projection)
Velocity	Tangential speed $v_{\text{circ}} = \omega A$ directed along tangent	Projection: $v(t) = \omega A \cos(\omega t + \phi_0)$
Acceleration	Centripetal acceleration $a_c = \omega^2 A$ directed towards center	Projection: $a(t) = -\omega^2 A \sin(\omega t + \phi_0) = -\omega^2 x$

Convention Note: In Indian JEE convention, the reference circle is typically traced counterclockwise. If the projection is taken on the $x$-axis from the vertical ($y$) reference axis, $x = A \sin(\omega t + \phi_0)$. If projected onto the $x$-axis from the horizontal axis, $x = A \cos(\omega t + \phi_0)$. Always verify the projection axis and initial direction of motion.

3. Graphical Representation: The Harmonic Curves ($x-t, v-t, a-t$)

Assuming the baseline initial condition of release from positive extreme ($x = A \cos \omega t$):

Kinematic Quantity	Analytical Equation	Amplitude Value	Relative Phase Lead
Displacement $x(t)$	$x(t) = A \cos(\omega t)$	$A$	$0$ (Reference)
Velocity $v(t)$	$v(t) = -\omega A \sin(\omega t) = \omega A \cos\left(\omega t + \frac{\pi}{2}\right)$	$v_{\text{max}} = \omega A$	Leads displacement by $\mathbf{+\frac{\pi}{2}\text{ rad} \ (90^\circ)}$
Acceleration $a(t)$	$a(t) = -\omega^2 A \cos(\omega t) = \omega^2 A \cos(\omega t + \pi)$	$a_{\text{max}} = \omega^2 A$	Leads displacement by $\mathbf{+\pi\text{ rad} \ (180^\circ)}$

• Key Phase Insights:
  • Velocity leads displacement by $\pi/2$ rad: when displacement is zero (mean position), velocity is maximum; when displacement is extreme, velocity is zero.
  • Acceleration leads velocity by $\pi/2$ rad, and leads displacement by $\pi$ rad (completely in anti-phase). When displacement is positive maximum ($+A$), acceleration is negative maximum ($-\omega^2 A$).

4. Phase-Space and State-Plane Trajectories

• $a-x$ Graph: Using $a = -\omega^2 x$, the plot of acceleration versus displacement is a straight line passing through the origin with negative slope: $$\text{Slope} = -\omega^2 = -\frac{4\pi^2}{T^2}$$ The angle of the line with the positive $x$-axis is obtuse: $\theta = \tan^{-1}(-\omega^2)$. The negative slope is the direct signature of a linear restoring force.
• $v-x$ Graph (Phase-Space Ellipse): Using the energy relation: $$v = \pm \omega \sqrt{A^2 – x^2} \implies \frac{v^2}{\omega^2} = A^2 – x^2 \implies \mathbf{\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1}$$ This is the standard equation of an ellipse with semi-major axis $A$ and semi-minor axis $\omega A$ (or vice versa).
  • If normalized coordinates $(x, v/\omega)$ are plotted, the trajectory becomes a circle of radius $A$.
  • The trajectory is traversed in a clockwise sense (for $x>0$, $v>0 \implies x$ is increasing; for $x>0, v<0 \implies x$ is decreasing).

5. Superposition of Two Perpendicular SHMs (Lissajous Figures)

When a particle is simultaneously subjected to two mutually perpendicular SHMs of the same frequency $\omega$:

$$x = A_1 \sin(\omega t), \quad y = A_2 \sin(\omega t + \delta)$$

Eliminating $t$ yields the general equation of a conic section:

$$\mathbf{\frac{x^2}{A_1^2} – \frac{2xy}{A_1 A_2}\cos \delta + \frac{y^2}{A_2^2} = \sin^2 \delta}$$

Phase Difference ($\delta$)	Trajectory Equation	Resulting Curve & Sense
$\delta = 0$	$\left(\frac{x}{A_1} – \frac{y}{A_2}\right)^2 = 0 \implies y = \frac{A_2}{A_1}x$	Straight line in 1st & 3rd quadrants (positive slope)
$\delta = \pi$	$\left(\frac{x}{A_1} + \frac{y}{A_2}\right)^2 = 0 \implies y = -\frac{A_2}{A_1}x$	Straight line in 2nd & 4th quadrants (negative slope)
$\delta = \pi/2$	$\frac{x^2}{A_1^2} + \frac{y^2}{A_2^2} = 1$	Symmetrical Ellipse; Circle if $A_1 = A_2$ (Clockwise)
$\delta = 3\pi/2$	$\frac{x^2}{A_1^2} + \frac{y^2}{A_2^2} = 1$	Symmetrical Ellipse; Circle if $A_1 = A_2$ (Counterclockwise)

Four Solved Examples (Graded Basic to Advanced)

Example 1: Basic Application — Identifying Phase Constant and Time to Reach Position

Problem Statement: A particle executes SHM of amplitude $A$ and time period $T$ along the $x$-axis about the origin $x=0$. Express its displacement in the form $x(t) = A \sin(\omega t + \phi_0)$ and determine its initial phase $\phi_0 \in [0, 2\pi)$ for each of the following initial conditions at $t = 0$:

(a) The particle is at $x = +A/2$ and moving towards the positive extreme.
(b) The particle is at $x = +A/2$ and moving towards the mean position.
(c) The particle is at $x = -A/\sqrt{2}$ and moving towards the negative extreme.

Solution:
The kinematic expressions are: $$x(t) = A \sin(\omega t + \phi_0), \quad v(t) = \omega A \cos(\omega t + \phi_0)$$ At $t = 0$: $x(0) = A \sin \phi_0$, and $v(0) = \omega A \cos \phi_0$.

Case (a): $x(0) = +A/2$ and $v(0) > 0$.
$$\sin \phi_0 = \frac{1}{2} \implies \phi_0 = \frac{\pi}{6} \text{ or } \frac{5\pi}{6}$$ Since $v(0) > 0$, $\cos \phi_0 > 0$. Thus, $\phi_0$ lies in the 1st quadrant: $$\mathbf{\phi_0 = \frac{\pi}{6} \text{ rad} \ (30^\circ)}$$

Case (b): $x(0) = +A/2$ and $v(0) < 0$.
Here $\sin \phi_0 = 1/2$, but the particle is moving in the $-x$ direction ($\cos \phi_0 < 0$). Thus $\phi_0$ lies in the 2nd quadrant: $$\mathbf{\phi_0 = \frac{5\pi}{6} \text{ rad} \ (150^\circ)}$$

Case (c): $x(0) = -A/\sqrt{2}$ and $v(0) < 0$.
$$\sin \phi_0 = -\frac{1}{\sqrt{2}} \implies \phi_0 = \frac{5\pi}{4} \text{ or } \frac{7\pi}{4}$$ Since it is moving towards the negative extreme, velocity is negative ($v(0) < 0 \implies \cos \phi_0 < 0$). This places the phasor in the 3rd quadrant: $$\mathbf{\phi_0 = \frac{5\pi}{4} \text{ rad} \ (225^\circ) \quad \left[\text{or } -\frac{3\pi}{4}\text{ rad}\right]}$$

Example 2: Intermediate Application — Two Particles Crossing with Equal Amplitude

Problem Statement: Two identical particles execute simple harmonic motion along the same straight line about the same mean position with identical amplitude $A$ and time period $T$. At a particular instant, they cross each other while moving in opposite directions at a displacement of $x = +A/2$.

(a) Determine the constant phase difference $\Delta \phi$ between them.
(b) Find the time elapsed after this crossing before they cross each other again.
(c) Find the maximum separation distance between the two particles during the motion.

Solution:
(a) Phase Difference via Reference Phasors:
Both particles are at position $x = +A/2$. Let Particle 1 be moving in the $+x$ direction and Particle 2 in the $-x$ direction. On the reference circle: $$\phi_1 = \sin^{-1}(1/2) \text{ with } \cos \phi_1 > 0 \implies \phi_1 = \frac{\pi}{6} = 30^\circ$$ $$\phi_2 = \sin^{-1}(1/2) \text{ with } \cos \phi_2 < 0 \implies \phi_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6} = 150^\circ$$ The phase difference between them is: $$\mathbf{\Delta \phi = \phi_2 - \phi_1 = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \text{ rad} \ (120^\circ)}$$

(b) Time to Next Crossing:
They cross each other when their displacements are equal: $x_1(t) = x_2(t)$.
$$A \sin(\omega t + \phi_1) = A \sin(\omega t + \phi_2)$$ $$\sin(\omega t + \phi_1) – \sin(\omega t + \phi_2) = 0 \implies 2 \cos\left(\omega t + \frac{\phi_1 + \phi_2}{2}\right) \sin\left(\frac{\phi_1 – \phi_2}{2}\right) = 0$$ Since $\Delta \phi = \phi_2 – \phi_1 = 2\pi/3 \neq 0$, we require: $$\cos\left(\omega t + \frac{\pi/6 + 5\pi/6}{2}\right) = 0 \implies \cos\left(\omega t + \frac{\pi}{2}\right) = 0 \implies -\sin(\omega t) = 0$$ Thus, $\omega t = n\pi \implies t = \frac{n\pi}{\omega} = n \frac{T}{2}$.
For the very next crossing ($n = 1$): $$\mathbf{\Delta t = \frac{T}{2}}$$ At this instant, the position is $x = A \sin\left(\pi + \frac{\pi}{6}\right) = -A/2$. They cross symmetrically at $x = -A/2$.

(c) Maximum Separation:
The separation between them is: $$\Delta x(t) = x_1(t) – x_2(t) = A [\sin(\omega t + \phi_1) – \sin(\omega t + \phi_2)] = 2A \sin\left(\frac{\phi_1 – \phi_2}{2}\right) \cos\left(\omega t + \frac{\phi_1 + \phi_2}{2}\right)$$ The maximum amplitude of separation is: $$\Delta x_{\text{max}} = 2A \left|\sin\left(\frac{\Delta \phi}{2}\right)\right| = 2A \sin\left(\frac{2\pi/3}{2}\right) = 2A \sin\left(\frac{\pi}{3}\right) = 2A \left(\frac{\sqrt{3}}{2}\right) = \mathbf{\sqrt{3}A \approx 1.732 A}$$

Example 3: Advanced Analysis — Reconstructing SHM Kinematics from $v-x$ and $a-x$ Phase Plots

Problem Statement: The velocity-displacement $(v-x)$ plot of a particle executing simple harmonic motion about $x = 0$ is an ellipse described by: $$25 v^2 + 625 \pi^2 x^2 = 625 \pi^2$$ where $x$ is in meters and $v$ is in $\text{m/s}$.

(a) Determine the amplitude $A$, angular frequency $\omega$, time period $T$, and maximum acceleration $a_{\text{max}}$.
(b) Write the equation of the corresponding $a-x$ graph and calculate its slope.
(c) If at $t = 0$ the particle is at $x = +1.0\text{ m}$ moving in the $-x$ direction, find the first instant $t_1 > 0$ when its velocity has magnitude half of maximum velocity.

Solution:
(a) Standard Ellipse Comparison:
Divide the given equation by $625 \pi^2$: $$\frac{25 v^2}{625 \pi^2} + \frac{625 \pi^2 x^2}{625 \pi^2} = 1 \implies \mathbf{x^2 + \frac{v^2}{25\pi^2} = 1}$$ Comparing with the canonical phase-space ellipse equation $\frac{x^2}{A^2} + \frac{v^2}{(\omega A)^2} = 1$: $$A^2 = 1 \implies \mathbf{A = 1.0\text{ m}}$$ $$(\omega A)^2 = 25\pi^2 \implies \omega A = 5\pi\text{ m/s} \implies \omega(1.0) = 5\pi\text{ rad/s} \implies \mathbf{\omega = 5\pi\text{ rad/s}}$$ $$\mathbf{T = \frac{2\pi}{\omega} = \frac{2\pi}{5\pi} = 0.40\text{ s}}$$ $$a_{\text{max}} = \omega^2 A = (5\pi)^2 (1.0) = \mathbf{25\pi^2 \approx 246.7\text{ m/s}^2}$$

(b) $a-x$ Graph Equation & Slope:
The relation between acceleration and displacement in SHM is: $$a = -\omega^2 x = -(5\pi)^2 x = \mathbf{-25\pi^2 x}$$ $$\mathbf{\text{Slope} = -\omega^2 = -25\pi^2 \approx -246.7\text{ s}^{-2}}$$ The graph is a straight line passing through the origin with slope $-25\pi^2$.

(c) Time Instant $t_1$:
At $t = 0$: $x(0) = +A = +1.0\text{ m}$, which means the particle is at the positive extreme! Thus: $$x(t) = A \cos(\omega t) = 1.0 \cos(5\pi t)$$ The velocity is: $$v(t) = -\omega A \sin(5\pi t) = -5\pi \sin(5\pi t)$$ We require $|v(t)| = \frac{1}{2} v_{\text{max}} = \frac{5\pi}{2}$: $$|-5\pi \sin(5\pi t)| = \frac{5\pi}{2} \implies |\sin(5\pi t)| = \frac{1}{2}$$ For the first instant $t_1 > 0$, the angle must be in the first quadrant: $$5\pi t_1 = \frac{\pi}{6} \implies \mathbf{t_1 = \frac{1}{30}\text{ s} \approx 0.0333\text{ s}}$$

Example 4: Advanced Application — Lissajous Trajectory and Elliptical Polarization

Problem Statement: A particle of mass $m = 0.2\text{ kg}$ is subjected to two mutually orthogonal harmonic force fields in the $x-y$ plane such that its motion is governed by: $$x(t) = A \sin(\omega t), \quad y(t) = A \sin\left(\omega t + \frac{\pi}{4}\right)$$ where $A = 10\text{ cm}$ and $\omega = 10\text{ rad/s}$.

(a) Derive the Cartesian trajectory equation $f(x, y) = 0$ traced by the particle and identify the geometric shape.
(b) Determine the coordinates of the points where the path intersects the line $y = x$.
(c) Calculate the kinetic energy of the particle when it passes through the origin $(x = 0)$.

Solution:
(a) Deriving the Trajectory:
Using the Lissajous formula for phase difference $\delta = \pi/4$: $$\frac{x^2}{A^2} – \frac{2xy}{A^2}\cos\left(\frac{\pi}{4}\right) + \frac{y^2}{A^2} = \sin^2\left(\frac{\pi}{4}\right)$$ Since $\cos(\pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}$, we have: $$\frac{x^2}{A^2} – \frac{\sqrt{2}xy}{A^2} + \frac{y^2}{A^2} = \frac{1}{2}$$ $$\mathbf{x^2 – \sqrt{2}xy + y^2 = \frac{A^2}{2}}$$ Substituting $A = 10\text{ cm} \implies A^2/2 = 50\text{ cm}^2$: $$\mathbf{x^2 – \sqrt{2}xy + y^2 = 50}$$ This represents an oblique ellipse whose major axis is oriented along the line $y = x$ ($\theta = 45^\circ$).

(b) Intersection with $y = x$:
Substitute $y = x$ into the trajectory equation: $$x^2 – \sqrt{2}x^2 + x^2 = 50 \implies (2 – \sqrt{2})x^2 = 50$$ $$x^2 = \frac{50}{2 – \sqrt{2}} = \frac{50(2 + \sqrt{2})}{4 – 2} = 25(2 + \sqrt{2}) \approx 25(3.4142) = 85.355\text{ cm}^2$$ $$x = \pm 5\sqrt{2 + \sqrt{2}} \approx \pm 9.24\text{ cm}$$ The points of intersection are: $$\mathbf{(+9.24\text{ cm}, +9.24\text{ cm}) \quad \text{and} \quad (-9.24\text{ cm}, -9.24\text{ cm})}$$

(c) Kinetic Energy at $x = 0$:
When $x(t) = 0$: $$A \sin(\omega t) = 0 \implies \omega t = 0 \text{ or } \pi$$ At $\omega t = 0$: $$v_x = \frac{dx}{dt} = \omega A \cos(0) = \omega A = (10\text{ rad/s})(0.10\text{ m}) = 1.0\text{ m/s}$$ $$v_y = \frac{dy}{dt} = \omega A \cos\left(0 + \frac{\pi}{4}\right) = \frac{\omega A}{\sqrt{2}} = \frac{1.0}{\sqrt{2}}\text{ m/s}$$ Total velocity squared is: $$v^2 = v_x^2 + v_y^2 = (1.0)^2 + \left(\frac{1.0}{\sqrt{2}}\right)^2 = 1.0 + 0.50 = 1.50\text{ m}^2/\text{s}^2$$ The kinetic energy is: $$K = \frac{1}{2}m v^2 = \frac{1}{2}(0.2\text{ kg})(1.50\text{ m}^2/\text{s}^2) = \mathbf{0.15\text{ J} = 150\text{ mJ}}$$

Practice Worksheet: 10 Target JEE Problems

Test your mastery of phase, phasor circles, and SHM graphs. Attempt all 10 problems before consulting the detailed solutions at the end.

Section I: Single Correct Choice Questions (Q1–Q5)

Q1. A body executes SHM with an amplitude $A$. At what displacement from the mean position is the phase of the motion equal to $45^\circ$ ($\pi/4$ rad), assuming displacement is zero at $t = 0$ with positive initial velocity?
(A) $A/2$
(B) $A/\sqrt{2}$
(C) $\sqrt{3}A/2$
(D) $A/4$

Q2. The displacement-time graph of a particle executing SHM is given by $x = A \sin(\omega t)$. The phase difference between the velocity $v(t)$ and the acceleration $a(t)$ of the particle is:
(A) $0$
(B) $\pi/4$
(C) $\pi/2$
(D) $\pi$

Q3. The acceleration $a$ versus displacement $x$ graph of a particle performing SHM is a straight line making an angle of $135^\circ$ with the positive $x$-axis. The time period of oscillation of the particle is:
(A) $2\pi\text{ s}$
(B) $\pi\text{ s}$
(C) $2\text{ s}$
(D) $4\pi\text{ s}$

Q4. Two simple harmonic motions are represented by $x_1 = 5 \sin(2\pi t + \pi/4)$ and $x_2 = 5\sqrt{2}(\sin 2\pi t + \cos 2\pi t)$. The ratio of the amplitude of the second motion to the first motion is:
(A) $1$
(B) $\sqrt{2}$
(C) $2$
(D) $1/\sqrt{2}$

Q5. A particle moves in the $x-y$ plane according to the equations $x = A \sin \omega t$ and $y = A \cos \omega t$. The path of the particle and the magnitude of its velocity are, respectively:
(A) Elliptical with speed $\omega A \sqrt{2}$
(B) Circular with constant speed $\omega A$
(C) Straight line with variable speed
(D) Parabolic with speed $\omega A$

Section II: One or More Than One Correct Questions (Q6–Q8)

Q6. A particle executes linear SHM along the $x$-axis with time period $T$ and amplitude $A$. Which of the following statements is/are correct regarding the phase and kinematics of the motion?
(A) The phase difference between two points where the particle has the same speed is either $0$ or $\pi – 2\phi$.
(B) When the kinetic energy and potential energy of the oscillator are equal, the phase of the motion is an odd multiple of $\pi/4$ (measured from mean position).
(C) The phase-space plot in the $(x, v/\omega)$ coordinate system is a circle of radius $A$ traversed clockwise.
(D) Acceleration always lags behind displacement by a phase of $\pi$ radians.

Q7. Two particles $P_1$ and $P_2$ execute SHM of the same amplitude $A$ and same angular frequency $\omega$ along parallel lines close to each other. When $P_1$ is at displacement $+A/\sqrt{2}$ moving towards the mean position, $P_2$ is at displacement $-A/2$ moving away from the mean position. Which of the following is/are correct?
(A) The phase constant of $P_1$ (relative to standard sine form) is $3\pi/4$.
(B) The phase constant of $P_2$ is $7\pi/6$.
(C) The phase difference between $P_1$ and $P_2$ can be $5\pi/12$ radians.
(D) The maximum separation between them along the line of motion is $A \sqrt{2 – \sqrt{3/2}}$.

Q8. For a particle in SHM, let $x, v,$ and $a$ represent displacement, velocity, and acceleration at time $t$. Which of the following graphs is/are mathematically correct?
(A) The $v^2$ versus $x^2$ graph is a straight line with negative slope $-\omega^2$.
(B) The $a^2$ versus $v^2$ graph is a straight line with negative slope $-\omega^2$.
(C) The $a$ versus $t$ curve is inverted (phase shifted by $\pi$) with respect to the $x$ versus $t$ curve.
(D) The total mechanical energy $E$ versus $x$ graph is a horizontal straight line independent of $x$.

Section III: Numerical / Integer Type Questions (Q9–Q10)

Q9. Two particles undergo SHM along the same straight line with the same amplitude $A = 12\text{ cm}$ and the same frequency. If the maximum separation between them during their motion is equal to the amplitude $A = 12\text{ cm}$, what is the phase difference between them in degrees? (Express answer as an integer in degrees $\le 180^\circ$).

Q10. The acceleration-displacement relation of a simple harmonic oscillator of mass $m = 0.5\text{ kg}$ is found from an experimental plot to be $a = -16 x$, where $x$ is in meters and $a$ is in $\text{m/s}^2$. If the maximum displacement of the oscillator is $5.0\text{ cm}$, find the maximum linear speed of the oscillator in $\text{cm/s}$.

Comprehensive Solutions & Answer Key

Answer Key

Question No.	Correct Answer	Key Concept / Formula Tested
Q1	(B)	$x = A \sin(\pi/4) = A/\sqrt{2}$
Q2	(C)	Phase lead of acceleration over velocity is $\pi/2$
Q3	(A)	$\text{Slope} = \tan(135^\circ) = -\omega^2 = -1 \implies T = 2\pi\text{ s}$
Q4	(C)	$A_2 = 5\sqrt{2}\sqrt{1^2 + 1^2} = 10 \implies A_2/A_1 = 10/5 = 2$
Q5	(B)	$x^2 + y^2 = A^2$, uniform speed $v = \omega A$
Q6	(B, C)	$U = K$ at $x = \pm A/\sqrt{2} \implies \phi = (2n+1)\pi/4$; phase circle
Q7	(A, B, C)	Reference circle phasors: $\phi_1 = 3\pi/4, \phi_2 = 7\pi/6 \implies \Delta \phi = 5\pi/12$
Q8	(A, B, C, D)	Kinematic relations: $v^2 = \omega^2(A^2 – x^2)$, $a^2 = \omega^2(v_{\text{max}}^2 – v^2)$, $E = \text{const}$
Q9	60	$S_{\text{max}} = 2A \sin(\Delta \phi/2) = A \implies \Delta \phi = 60^\circ$
Q10	20	$\omega^2 = 16 \implies \omega = 4\text{ rad/s} \implies v_{\text{max}} = \omega A = 20\text{ cm/s}$

Detailed Step-by-Step Solutions

Solution to Q1:
The displacement is given by $x(t) = A \sin \theta$, where $\theta = \omega t + \phi_0$ is the instantaneous phase.
Given $\theta = 45^\circ = \pi/4$ rad:
$$x = A \sin\left(\frac{\pi}{4}\right) = \mathbf{\frac{A}{\sqrt{2}}}$$ Hence, the correct choice is (B).

Solution to Q2:
Given $x(t) = A \sin(\omega t)$: $$v(t) = \frac{dx}{dt} = \omega A \cos(\omega t) = \omega A \sin\left(\omega t + \frac{\pi}{2}\right)$$ $$a(t) = \frac{dv}{dt} = -\omega^2 A \sin(\omega t) = \omega^2 A \sin(\omega t + \pi) = \omega^2 A \cos\left(\omega t + \frac{\pi}{2} + \frac{\pi}{2}\right)$$ The phase of $v(t)$ is $\phi_v = \omega t + \pi/2$. The phase of $a(t)$ is $\phi_a = \omega t + \pi$.
The phase difference between acceleration and velocity is: $$\phi_a – \phi_v = (\omega t + \pi) – \left(\omega t + \frac{\pi}{2}\right) = \mathbf{\frac{\pi}{2}\text{ rad}}$$ Hence, acceleration leads velocity by $\pi/2$ radians. Correct choice is (C).

Solution to Q3:
In SHM, the relation between acceleration and displacement is $a = -\omega^2 x$.
In the $a$ versus $x$ graph, this is a straight line $y = mx$ with slope $m = -\omega^2$.
Given the line makes an angle $\theta = 135^\circ$ with the positive $x$-axis: $$\text{Slope } m = \tan(135^\circ) = -1$$ $$-\omega^2 = -1 \implies \omega^2 = 1 \implies \omega = 1\text{ rad/s}$$ The time period of oscillation is: $$T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = \mathbf{2\pi\text{ s}}$$ Hence, the correct choice is (A).

Solution to Q4:
For the first SHM: $x_1 = 5 \sin(2\pi t + \pi/4)$, the amplitude is directly $A_1 = 5$.
For the second SHM: $$x_2 = 5\sqrt{2} \sin(2\pi t) + 5\sqrt{2} \cos(2\pi t)$$ This is the superposition of two perpendicular phasors of amplitude $B = 5\sqrt{2}$: $$A_2 = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2} = \sqrt{50 + 50} = \sqrt{100} = 10$$ The ratio of amplitudes is: $$\frac{A_2}{A_1} = \frac{10}{5} = \mathbf{2}$$ Hence, the correct choice is (C).

Solution to Q5:
We have $x = A \sin \omega t$ and $y = A \cos \omega t$.
Squaring and adding both equations: $$x^2 + y^2 = A^2 \sin^2 \omega t + A^2 \cos^2 \omega t = A^2 (\sin^2 \omega t + \cos^2 \omega t) = A^2$$ This is the equation of a circle of radius $A$ centered at the origin.
Velocity components: $$v_x = \frac{dx}{dt} = \omega A \cos \omega t, \quad v_y = \frac{dy}{dt} = -\omega A \sin \omega t$$ Total speed is: $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(\omega A)^2 \cos^2 \omega t + (\omega A)^2 \sin^2 \omega t} = \omega A \sqrt{\cos^2 \omega t + \sin^2 \omega t} = \mathbf{\omega A = \text{constant}}$$ Thus, the motion is uniform circular motion with constant speed $\omega A$. Correct choice is (B).

Solution to Q6:
Let us evaluate each statement:
(A) The speed is $v = \omega \sqrt{A^2 – x^2}$. Two points have the same speed if they have the same $|x|$. For a given $x$, the two directions of motion correspond to phases $\phi$ and $\pi – \phi$, so the phase difference is $\pi – 2\phi$. For symmetrical points $+x$ and $-x$, the phase difference is $\pi$. Thus (A) is an incomplete characterization.
(B) $U = \frac{1}{2} k x^2$ and $K = \frac{1}{2} k (A^2 – x^2)$. Setting $U = K \implies x^2 = A^2/2 \implies x = \pm A/\sqrt{2}$.
In $x = A \sin \theta$, $\sin \theta = \pm 1/\sqrt{2} \implies \theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$, which are odd multiples of $\pi/4$. Thus (B) is correct.
(C) In coordinates $(X, Y) = (x, v/\omega)$, we have $X = A \sin(\omega t + \phi_0)$ and $Y = A \cos(\omega t + \phi_0)$. Then $X^2 + Y^2 = A^2$, which is a circle of radius $A$. Since $Y = v/\omega = dX/d(\omega t)$, when $X > 0$, $Y$ can be positive (moving right, $X$ increasing) and then becomes negative (moving left, $X$ decreasing), which traces a clockwise circle. Thus (C) is correct.
(D) $a = -\omega^2 x = \omega^2 x \cos(\pi) = \omega^2 A \sin(\omega t + \pi)$. Acceleration leads displacement by $\pi$ radians (or leads/lags symmetrically by $\pi$). In standard phase convention, $a(t)$ leads $x(t)$ by $+\pi$. Claiming it ‘always lags’ is incorrect in rigorous phase lead formulation. Hence, correct choices are (B, C).

Solution to Q7:
For Particle $P_1$: $x_1 = +A/\sqrt{2}$, moving towards the mean position ($v_1 < 0$).
$$\sin \phi_1 = \frac{1}{\sqrt{2}} \text{ with } \cos \phi_1 < 0 \implies \phi_1 = \pi - \frac{\pi}{4} = \mathbf{\frac{3\pi}{4}\text{ rad}}$$ Thus statement (A) is correct.

For Particle $P_2$: $x_2 = -A/2$, moving away from the mean position (towards $-A \implies v_2 < 0$).
$$\sin \phi_2 = -\frac{1}{2} \text{ with } \cos \phi_2 < 0 \implies \phi_2 = \pi + \frac{\pi}{6} = \mathbf{\frac{7\pi}{6}\text{ rad}}$$ Thus statement (B) is correct.

Phase difference between them: $$\Delta \phi = \phi_2 – \phi_1 = \frac{7\pi}{6} – \frac{3\pi}{4} = \frac{14\pi – 9\pi}{12} = \mathbf{\frac{5\pi}{12}\text{ rad} \ (75^\circ)}$$ Thus statement (C) is correct.

Maximum separation: $$\Delta x_{\text{max}} = 2A \sin\left(\frac{\Delta \phi}{2}\right) = 2A \sin\left(\frac{5\pi}{24}\right) \neq A\sqrt{2 – \sqrt{3/2}}$$ Thus (D) is incorrect. Correct choices are (A, B, C).

Solution to Q8:
Let us analyze each option:
(A) We know $v^2 = \omega^2 (A^2 – x^2) \implies v^2 = -\omega^2 x^2 + \omega^2 A^2$. This is a linear relation between $v^2$ (as $y$) and $x^2$ (as $x$) with negative slope $-\omega^2$. Thus (A) is correct.
(B) Acceleration magnitude is $a = -\omega^2 x \implies x^2 = a^2/\omega^4$.
Substitute into velocity: $v^2 = \omega^2 A^2 – \omega^2 \left(\frac{a^2}{\omega^4}\right) = v_{\text{max}}^2 – \frac{a^2}{\omega^2} \implies a^2 = -\omega^2 v^2 + \omega^2 v_{\text{max}}^2$. This is a straight line between $a^2$ and $v^2$ with negative slope $-\omega^2$. Thus (B) is correct.
(C) $a(t) = -\omega^2 x(t) = \omega^2 A \sin(\omega t + \pi)$. The acceleration waveform is an inverted replica of the displacement waveform with a phase shift of $\pi$. Thus (C) is correct.
(D) Total mechanical energy $E = \frac{1}{2} k A^2 = \text{constant}$, which is a horizontal line on an energy versus displacement graph. Thus (D) is correct.
Hence, all options (A, B, C, D) are correct.

Solution to Q9:
The separation between two particles of identical amplitude $A$ and phase difference $\Delta \phi$ is given by: $$\Delta x(t) = x_1(t) – x_2(t) = 2A \sin\left(\frac{\Delta \phi}{2}\right) \cos\left(\omega t + \frac{\phi_1 + \phi_2}{2}\right)$$ The maximum value of separation is: $$S_{\text{max}} = 2A \sin\left(\frac{\Delta \phi}{2}\right)$$ We are given that $S_{\text{max}} = A$ and $A = 12\text{ cm}$: $$2A \sin\left(\frac{\Delta \phi}{2}\right) = A \implies \sin\left(\frac{\Delta \phi}{2}\right) = \frac{1}{2}$$ $$\frac{\Delta \phi}{2} = 30^\circ \implies \mathbf{\Delta \phi = 60^\circ}$$ Answer: 60

Solution to Q10:
The acceleration relation is given as: $$a = -16 x$$ Comparing with the fundamental equation of SHM $a = -\omega^2 x$: $$\omega^2 = 16 \implies \mathbf{\omega = 4.0\text{ rad/s}}$$ Given amplitude $A = 5.0\text{ cm}$, the maximum velocity is: $$v_{\text{max}} = \omega A = (4.0\text{ rad/s})(5.0\text{ cm}) = \mathbf{20\text{ cm/s}}$$ Answer: 20