Step 1: Identifying the General Equation of SHM

The standard equation for Simple Harmonic Motion (SHM) is given by:

$x(t) = A \sin(\omega t + \phi)$

Given the equation: $x = -2 \sin(3t + \pi/3)$ cm

To align with the reference circle convention where $x = A \sin(\omega t + \theta)$, we use the trigonometric identity $-\sin(\theta) = \sin(\theta + \pi)$ [Per the Unit Circle Symmetry Principle]:

$x = 2 \sin(3t + \pi/3 + \pi)$

$x = 2 \sin(3t + 4\pi/3)$

Step 2: Extraction of Parameters

• Amplitude ($A$): $2 \text{ cm}$ [Representing the radius of the reference circle]
• Angular Speed ($\omega$): $3 \text{ rad/s}$
• Initial Phase ($\phi$): $4\pi/3 \text{ rad}$ (equivalent to $240^\circ$)

Step 3: Reference Circle Visualization

In the reference circle method, the projection of a particle rotating with constant angular velocity $\omega$ on the $x$-axis represents the SHM. The radius of the circle is the amplitude $A$.

Step 4: Analytical Justification of Initial Position

At $t = 0$, the reference particle is at an angle $\theta = 4\pi/3$ radians relative to the positive $x$-axis. Using coordinate geometry: $x_0 = A \cos(4\pi/3) = 2 \cos(240^\circ) = 2(-0.5) = -1 \text{ cm}$ $y_0 = A \sin(4\pi/3) = 2 \sin(240^\circ) = 2(-\sqrt{3}/2) = -\sqrt{3} \approx -1.732 \text{ cm}$

This confirms the position of the particle at the start of the observation period in the 3rd quadrant.

Final Solution: The reference circle has a radius of $2 \text{ cm}$, an angular speed of $3 \text{ rad/s}$, and the particle starts at an angular position of $4\pi/3$ radians ($240^\circ$) measured counter-clockwise from the positive $x$-axis.

Given Variables & Initial Setup

In a linear Simple Harmonic Motion (SHM) system occurring between two points $A$ and $B$, the total distance $AB = 10 \text{ cm}$. Let the equilibrium position (mean position) be $O$.

• Total path length = $2A = 10 \text{ cm}$, where $A$ is the amplitude. Therefore, $A = 5 \text{ cm}$.
• Mean position ($O$) is at $x = 0$.
• Point $A$ is at $x = -5 \text{ cm}$.
• Point $B$ is at $x = +5 \text{ cm}$.
• Direction $A \to B$ is defined as the positive direction ($+x$).

Visual Representation of the Oscillatory Path

Step 1: Determining Velocity ($v$) at Point A

At the extreme positions of SHM, the particle momentarily comes to rest before changing direction. Since point $A$ is an extreme position, the instantaneous velocity of the particle is zero.

$v = 0$ [Per the definition of extreme points in SHM where $v = \omega\sqrt{A^2 - x^2}$ and $x = \pm A$].

Step 2: Determining Acceleration ($a$) at Point A

The acceleration of a particle in SHM is given by $a = -\omega^2 x$.

• At point $A$, the displacement $x = -5 \text{ cm}$.
• Substituting the value: $a = -\omega^2 (-5 \text{ cm}) = +5\omega^2 \text{ cm/s}^2$.
• Since the result is positive, the acceleration vector points in the positive direction ($A \to B$).

[Justification: According to the restoring force principle, acceleration is always directed towards the equilibrium position $O$. Since $A$ is to the left of $O$, the force/acceleration must act towards the right].

Step 3: Determining Force ($F$) at Point A

According to Newton's Second Law of Motion ($F = ma$):

• Since mass ($m$) is a positive scalar, the direction of the force is identical to the direction of acceleration.
• As determined in Step 2, the acceleration is positive. Therefore, the force is positive.

[Justification: $F = -k x$. With $x = -5$, $F = -k(-5) = +5k$, which is positive].

Summary of Signs at Point A

Final Solution: At point A, the velocity is 0, the acceleration is positive, and the force is positive.

Theoretical Foundation & Definition

Periodic motion is defined as a motion that repeats itself at regular intervals of time, known as the period ($T$). Mathematically, a physical quantity $f(t)$ representing the motion must satisfy the condition: $f(t) = f(t + T)$, where $T > 0$ is the smallest constant interval.

Step 1: Analysis of Rotational Dynamics

A hydrogen molecule ($H_2$) consists of two hydrogen atoms linked by a covalent bond, essentially behaving as a rigid rotor when rotating about its center of mass. In a stable, isolated environment, the molecule possesses a fixed angular momentum and rotates with a constant angular velocity ($\omega$).

The position of any point on the molecule can be described using angular coordinates in polar or spherical systems. The angular displacement $\theta$ as a function of time $t$ is given by: $\theta(t) = \omega t + \theta_0$

Step 2: Testing for Periodicity

We check if the orientation of the molecule repeats at regular intervals. Let the orientation be defined by the angle $\theta$. The motion is periodic if $\theta(t + T) = \theta(t) + 2\pi n$ for some integer $n$.

• $\omega(t + T) + \theta_0 = \omega t + \theta_0 + 2\pi$ [Per the definition of circular periodicity]
• $\omega t + \omega T + \theta_0 = \omega t + \theta_0 + 2\pi$
• $\omega T = 2\pi$
• $T = \frac{2\pi}{\omega}$

Since $\omega$ is constant for a free rotating molecule, $T$ is a constant, finite value.

Visual Representation of Rotational Periodic Motion

Conclusion and Justification

The rotation of a hydrogen molecule about its center of mass is a classic example of uniform circular motion. Because the angular velocity remains constant in the absence of external torques, the molecule returns to the same spatial orientation after every time interval $T = \frac{2\pi}{\omega}$. Therefore, the motion is strictly periodic.

Final Solution: Yes, the rotation of a hydrogen molecule about its center of mass represents a periodic motion, as it repeats its spatial orientation at a constant time interval $T = \frac{2\pi}{\omega}$.

Step 1: Establishing the Equations of Motion

For a particle undergoing Simple Harmonic Motion (SHM) along a straight line, the displacement $x(t)$ as a function of time is given by:

$x(t) = A \sin(\omega t + \phi)$

[Per the definition of SHM, where $A$ is amplitude, $\omega$ is angular frequency, and $\phi$ is the phase constant].

The instantaneous velocity $v(t)$ is the first derivative of displacement with respect to time:

$v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$

Step 2: Defining Kinetic and Potential Energy

The instantaneous Kinetic Energy ($K$) is defined as:

$K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t + \phi)$

[Using the relation $k = m\omega^2$, where $k$ is the force constant, we can write $K = \frac{1}{2} k A^2 \cos^2(\omega t + \phi)$].

The instantaneous Potential Energy ($U$) is defined as:

$U = \frac{1}{2}kx^2 = \frac{1}{2} k A^2 \sin^2(\omega t + \phi)$

Step 3: Calculating Average Energy Over One Period ($T$)

The average value of a function $f(t)$ over a period $T$ is given by $\langle f \rangle = \frac{1}{T} \int_{0}^{T} f(t) dt$.

For Kinetic Energy:

$\langle K \rangle = \frac{1}{T} \int_{0}^{T} \frac{1}{2} k A^2 \cos^2(\omega t + \phi) dt$

Using the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$\langle K \rangle = \frac{kA^2}{2T} \int_{0}^{T} \frac{1 + \cos(2(\omega t + \phi))}{2} dt$

Since the integral of $\cos(2\omega t + 2\phi)$ over a full period $T = \frac{2\pi}{\omega}$ is zero:

$\langle K \rangle = \frac{kA^2}{4T} [t]_{0}^{T} = \frac{kA^2}{4T} \cdot T = \frac{1}{4}kA^2$

Step 4: Calculation for Potential Energy

Similarly, for Potential Energy:

$\langle U \rangle = \frac{1}{T} \int_{0}^{T} \frac{1}{2} k A^2 \sin^2(\omega t + \phi) dt$

Using the identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$:

$\langle U \rangle = \frac{kA^2}{2T} \int_{0}^{T} \frac{1 - \cos(2(\omega t + \phi))}{2} dt$

$\langle U \rangle = \frac{kA^2}{4T} [t]_{0}^{T} = \frac{1}{4}kA^2$

Visual Representation of Energy Distribution

Conclusion

Comparing the derived averages:

Final Solution: Since $\langle K \rangle = \frac{1}{4}kA^2$ and $\langle U \rangle = \frac{1}{4}kA^2$, it is proven that $\langle K \rangle = \langle U \rangle$ for a particle in linear SHM.

1. Mathematical Representation & Analysis

The given function is defined as:

$f(t) = \cos(\omega t) + \cos(3\omega t) + \cos(5\omega t)$

2. Testing for Periodicity

A function $f(t)$ is periodic if there exists a smallest positive constant $T$ such that $f(t + T) = f(t)$.

The individual terms have periods:

• $T_1 = \frac{2\pi}{\omega}$
• $T_2 = \frac{2\pi}{3\omega} = \frac{T_1}{3}$
• $T_3 = \frac{2\pi}{5\omega} = \frac{T_1}{5}$

[By the Principle of Superposition of Periodic Functions]: The sum of periodic functions is periodic if the ratio of their periods is a rational number. Here, the ratios are $T_1:T_2:T_3 = 1 : \frac{1}{3} : \frac{1}{5} = 15 : 5 : 3$. Since these are rational, the function is periodic.

3. Determining the Fundamental Period ($T$)

The fundamental period $T$ is the Least Common Multiple (LCM) of the individual periods $T_1, T_2,$ and $T_3$.

$T = \text{LCM}\left( \frac{2\pi}{\omega}, \frac{2\pi}{3\omega}, \frac{2\pi}{5\omega} \right)$

$T = \frac{2\pi}{\omega} \times \text{LCM}\left( 1, \frac{1}{3}, \frac{1}{5} \right)$

To find the LCM of fractions, we use the formula: $\text{LCM}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{GCD}(b, d)}$.

$T = \frac{2\pi}{\omega} \times \frac{\text{LCM}(1, 1, 1)}{\text{GCD}(1, 3, 5)} = \frac{2\pi}{\omega} \times \frac{1}{1} = \frac{2\pi}{\omega}$

4. Classification of Motion

Simple Harmonic Motion (SHM) must be representable as a single sinusoidal function of the form $A \cos(\omega t + \phi)$. Since $f(t)$ is a superposition of three distinct frequencies ($\omega, 3\omega, 5\omega$), it does not satisfy the definition of SHM.

5. Visual Representation (Superposition Principle)

Final Solution: The function represents a periodic but not simple harmonic motion. Its fundamental period is $T = \frac{2\pi}{\omega}$.

Given Variables & Initial Setup

To determine the maximum speed of the piston undergoing Simple Harmonic Motion (SHM), we extract the following parameters from the problem statement:

• Stroke length ($L$) = $1.0\text{ m}$
• Angular frequency ($\omega$) = $200\text{ rad/min}$

Step 1: Determining Amplitude ($A$)

The stroke is defined as the total distance covered by the piston in one full cycle, which is equivalent to twice the amplitude ($L = 2A$). [Per the definition of SHM kinematics]

$A = \frac{L}{2} = \frac{1.0\text{ m}}{2} = 0.5\text{ m}$

Step 2: Unit Conversion for Angular Frequency

To maintain consistency with SI units, the angular frequency must be converted from radians per minute to radians per second. [Per the International System of Units (SI) standards]

$\omega = 200\text{ rad/min} = \frac{200}{60}\text{ rad/s}$

$\omega = \frac{10}{3}\text{ rad/s} \approx 3.33\text{ rad/s}$

Step 3: Deriving Maximum Speed ($v_{max}$)

In SHM, the displacement is given by $x(t) = A \sin(\omega t + \phi)$. The velocity is the first derivative of displacement with respect to time: $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. The maximum speed occurs when the cosine term is unity. [By the Fundamental Theorem of Calculus]

$v_{max} = A \cdot \omega$

Step 4: Calculation

Substituting the values derived in Step 1 and Step 2:

$v_{max} = 0.5\text{ m} \times \frac{10}{3}\text{ rad/s}$

$v_{max} = \frac{5}{3}\text{ m/s}$

$v_{max} \approx 1.67\text{ m/s}$

Final Solution: The maximum speed of the piston is approximately $1.67\text{ m/s}$.

Step 1: Analyzing the Forces in the Non-Inertial Frame

To determine the time period of the pendulum, we must transition to the non-inertial frame of reference of the moving car. In this frame, the pendulum bob experiences two primary accelerations:

• Gravitational Acceleration ($g$): Acting vertically downwards.
• Centripetal Acceleration ($a_c$): Directed horizontally towards the center of the circular track. Given the speed $v$ and radius $R$, $a_c = \frac{v^2}{R}$.

As the car moves, the bob experiences a pseudo-force directed radially outward, equal to $M \frac{v^2}{R}$.

Step 2: Determining the Effective Acceleration ($g_{eff}$)

The equilibrium position of the pendulum is shifted such that the tension in the string balances the resultant of the gravitational force and the pseudo-force. The effective acceleration $g_{eff}$ is the vector sum of the acceleration due to gravity and the centripetal acceleration:

$\vec{g}_{eff} = \vec{g} + \vec{a}_c$

Since $\vec{g}$ (vertical) and $\vec{a}_c$ (horizontal) are mutually perpendicular, the magnitude of the effective acceleration is calculated using the Pythagorean theorem [Per the Law of Vector Addition]:

$g_{eff} = \sqrt{g^2 + a_c^2}$

$g_{eff} = \sqrt{g^2 + \left( \frac{v^2}{R} \right)^2}$

Step 3: Calculating the Time Period

For a simple pendulum undergoing small oscillations, the time period $T$ is given by the standard formula [By Simple Harmonic Motion theory for a pendulum]:

$T = 2\pi \sqrt{\frac{l}{g_{eff}}}$

Substituting the expression for $g_{eff}$ derived in Step 2:

$T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + \left( \frac{v^2}{R} \right)^2}}}$

Step 4: Geometric Visualization

Step 5: Final Result

The time period depends on the effective gravity experienced by the bob in the rotating frame of reference. By substituting the components:

$T = 2\pi \left( \frac{l^2}{g^2 + \frac{v^4}{R^2}} \right)^{1/4}$

Final Solution: $T = 2\pi \sqrt{\frac{l}{(g^2 + v^4/R^2)^{1/2}}}$ seconds

Step 1: Identifying the Mathematical Form

The given function of time is $f(t) = 1 + \omega t + \omega^2 t^2$. This is a second-degree polynomial in $t$ [Reference: Algebraic Theory of Polynomials].

Step 2: Checking for Periodicity

A function $f(t)$ is defined as periodic if there exists a smallest positive constant $T$ (the period) such that $f(t + T) = f(t)$ for all $t$ [Definition of Periodicity].

Let us test the condition $f(t + T) = f(t)$:

$1 + \omega(t + T) + \omega^2(t + T)^2 = 1 + \omega t + \omega^2 t^2$

Expanding the left side:

$1 + \omega t + \omega T + \omega^2(t^2 + 2tT + T^2) = 1 + \omega t + \omega^2 t^2$

$1 + \omega t + \omega T + \omega^2 t^2 + 2\omega^2 tT + \omega^2 T^2 = 1 + \omega t + \omega^2 t^2$

Subtracting $1 + \omega t + \omega^2 t^2$ from both sides:

$\omega T + 2\omega^2 tT + \omega^2 T^2 = 0$

Step 3: Analytical Conclusion

For the equality to hold, the expression $\omega T + 2\omega^2 tT + \omega^2 T^2$ must equal zero for all values of $t$. However, this expression is a function of $t$ (specifically, it is linear in $t$ because of the term $2\omega^2 tT$).

Since $\omega$ is a non-zero positive constant, the coefficient of $t$, which is $2\omega^2 T$, must be zero for the entire expression to be independent of $t$. This implies $T = 0$.

However, by definition, the period $T$ of a periodic motion must be a positive constant ($T > 0$). Since no such non-zero $T$ exists, the function does not satisfy the requirements of periodic motion [Theorem of Periodic Functions].

Visual Representation of Non-Periodic Growth

Final Categorization

As the function $f(t) = 1 + \omega t + \omega^2 t^2$ grows indefinitely as $t \to \infty$ and does not repeat its values at any regular interval, it is classified as non-periodic.

Final Solution: The function $1 + \omega t + \omega^2 t^2$ represents (c) non-periodic motion.

Step 1: Given Variables & Theoretical Foundation

To determine the torsional spring constant $\alpha$, we utilize the dynamics of a torsional pendulum. The system consists of a circular disc suspended by a wire, undergoing angular simple harmonic motion.

• Mass of the disc ($M$): $10\text{ kg}$
• Radius of the disc ($R$): $15\text{ cm} = 0.15\text{ m}$
• Period of oscillation ($T$): $1.5\text{ s}$
• Restoring torque relation: $J = -\alpha \theta$

[By the Principle of Angular Harmonic Motion, the period $T$ is given by $T = 2\pi \sqrt{\frac{I}{\alpha}}$, where $I$ is the Moment of Inertia of the disc about the axis of rotation.]

Step 2: Calculation of Moment of Inertia ($I$)

For a circular disc rotating about an axis passing through its centre and perpendicular to its plane, the moment of inertia is defined as:

$I = \frac{1}{2} M R^2$

Substituting the given values:

$I = \frac{1}{2} \times 10\text{ kg} \times (0.15\text{ m})^2$

$I = 5 \times 0.0225 = 0.1125\text{ kg}\cdot\text{m}^2$

Step 3: Deriving the Torsional Spring Constant ($\alpha$)

Rearranging the period formula $T = 2\pi \sqrt{\frac{I}{\alpha}}$ to solve for $\alpha$:

$\frac{T}{2\pi} = \sqrt{\frac{I}{\alpha}}$

$\frac{T^2}{4\pi^2} = \frac{I}{\alpha}$

$\alpha = \frac{4\pi^2 I}{T^2}$

Step 4: Final Substitution & Numerical Computation

Using $\pi^2 \approx 9.8696$:

$\alpha = \frac{4 \times 9.8696 \times 0.1125}{(1.5)^2}$

$\alpha = \frac{4.44132}{2.25}$

$\alpha \approx 1.9739\text{ N}\cdot\text{m/rad}$

Final Solution: The torsional spring constant of the wire is approximately $1.97\text{ N}\cdot\text{m/rad}$.

Step 1: Theoretical Foundation of SHM Projection

A particle undergoing uniform circular motion at radius $A$ and angular frequency $\omega$ has a position vector $\vec{r}(t)$. The projection of this motion on the $x$-axis follows the equation of Simple Harmonic Motion (SHM):

$x(t) = A \cos(\omega t + \phi)$

Where:

• $A$ = Amplitude (Radius of the circle)
• $\omega = \frac{2\pi}{T}$ = Angular frequency ($T$ is the period of revolution)
• $\phi$ = Initial phase angle (the angle the position vector makes with the positive $x$-axis at $t=0$)

Step 2: Analysis of Figure (a)

Given Parameters:

• Radius $A = 3\text{ cm}$
• Period $T = 2\text{ s} \implies \omega = \frac{2\pi}{2} = \pi \text{ rad/s}$
• Initial position at $t=0$: The vector is at $\theta = 45^\circ$ ($\frac{\pi}{4} \text{ rad}$) from the positive $x$-axis.
• Direction: Anti-clockwise (positive angular velocity).

Substituting into the standard SHM equation:

$x(t) = 3 \cos\left(\pi t + \frac{\pi}{4}\right)$

Step 3: Analysis of Figure (b)

Given Parameters:

• Radius $A = 2\text{ m}$
• Period $T = 4\text{ s} \implies \omega = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s}$
• Initial position at $t=0$: The vector is at $\theta = 90^\circ$ ($\frac{\pi}{2} \text{ rad}$) from the positive $x$-axis.
• Direction: Clockwise (negative angular velocity).

For clockwise motion, the angular displacement is $-\omega t$. Thus, the phase becomes $(\phi - \omega t)$:

$x(t) = 2 \cos\left(\frac{\pi}{2}t - \frac{\pi}{2}\right)$

Using the trigonometric identity $\cos(\theta - \frac{\pi}{2}) = \sin(\theta)$:

$x(t) = 2 \sin\left(\frac{\pi}{2}t\right)$

Visual Representation of Projection

Data Summary Table

Final Solution: The equations of motion for the $x$-projections are (a) $x(t) = 3 \cos(\pi t + \pi/4)$ and (b) $x(t) = 2 \sin(\pi t/2)$ in their respective units.

Step 1: Identifying the General Equation of Simple Harmonic Motion (SHM)

The displacement of a particle executing Simple Harmonic Motion at any time $t$ is given by the general displacement equation:

$x(t) = A \sin(\omega t + \phi)$

Where:

• $A$: Amplitude of oscillation (maximum displacement from mean position).
• $\omega$: Angular frequency of oscillation, defined as $\sqrt{\frac{k}{m}}$.
• $\phi$: Initial phase constant (phase angle at $t=0$).

Step 2: Applying Initial Conditions

According to the problem statement, at $t = 0$, the mass is at the mean position. By the coordinate system defined, the mean position corresponds to:

$x(0) = 0$

Step 3: Calculating the Phase Constant ($\phi$)

Substitute the initial condition into the general equation:

$0 = A \sin(\omega(0) + \phi)$

$0 = A \sin(\phi)$

Assuming the amplitude $A \neq 0$ for an oscillating system, we have:

$\sin(\phi) = 0 \implies \phi = 0$ or $\phi = \pi$

Taking the standard phase convention where the particle starts from the mean position moving in the positive direction, we set $\phi = 0$.

Step 4: Formulating the Displacement Function

Substituting $\phi = 0$ into the general displacement equation:

$x(t) = A \sin(\omega t + 0)$

$x(t) = A \sin(\omega t)$

Visual Representation of the Resulting Displacement Wave

Final Analytical Synthesis

The displacement of the mass as a function of time, starting from the mean position ($x=0$ at $t=0$), is governed by the sine function. The motion is purely periodic and passes through the origin at $t=0, \frac{\pi}{\omega}, \frac{2\pi}{\omega}, \dots$

Final Solution: $x(t) = A \sin(\omega t)$

Step 1: Establishing the Initial Equilibrium State

Let $P$ be the atmospheric pressure and $V$ be the volume of the air chamber. When the ball of mass $m$ is at equilibrium, the downward force due to its weight ($mg$) is balanced by the upward force due to the pressure difference. However, if the pressure inside the chamber is equal to the atmospheric pressure $P$, the ball rests at the opening. Let the pressure inside the chamber be $P$.

Step 2: Analysis of Isothermal Compression

Suppose the ball is pushed down by a small distance $x$ into the neck of area $a$. The decrease in volume is $\Delta V = ax$. [By the definition of isothermal processes, the product of pressure and volume remains constant].

The initial state is $(P, V)$. The final state after compression is $(P + \Delta P, V - \Delta V)$.

According to Boyle's Law: $PV = (P + \Delta P)(V - \Delta V)$

$PV = PV - P\Delta V + V\Delta P - \Delta P \Delta V$

Neglecting the small product term $\Delta P \Delta V$:

$P\Delta V = V\Delta P \implies \Delta P = \frac{P \Delta V}{V} = \frac{Pax}{V}$

Step 3: Calculating the Restoring Force

The excess pressure $\Delta P$ acts on the bottom of the ball, exerting an upward force $F$.

$F = -(\Delta P) \cdot a$ [The negative sign indicates the force is restorative, directed opposite to the displacement $x$].

Substituting the expression for $\Delta P$:

$F = -\left( \frac{Pax}{V} \right) a = -\left( \frac{Pa^2}{V} \right) x$

Step 4: Proving Simple Harmonic Motion (SHM)

According to Newton's Second Law, $F = ma = m \frac{d^2x}{dt^2}$.

$m \frac{d^2x}{dt^2} = -\left( \frac{Pa^2}{V} \right) x$

$\frac{d^2x}{dt^2} = -\left( \frac{Pa^2}{mV} \right) x$

This equation is of the form $\frac{d^2x}{dt^2} = -\omega^2 x$, which is the defining differential equation of SHM [Per the definition of linear restorative forces].

Step 5: Deriving the Time Period

Comparing the derived equation with the standard SHM equation, we identify the angular frequency $\omega$:

$\omega^2 = \frac{Pa^2}{mV} \implies \omega = \sqrt{\frac{Pa^2}{mV}}$

The time period $T$ is given by $T = \frac{2\pi}{\omega}$ [Relationship between angular frequency and periodic time]:

$T = 2\pi \sqrt{\frac{mV}{Pa^2}}$

Final Solution: The system executes SHM because the restoring force is directly proportional to displacement $x$ ($F \propto -x$). The time period of oscillation is $T = 2\pi \sqrt{\frac{mV}{Pa^2}}$.

Step 1: Analyzing the Given Function

The given function of time is:

$f(t) = 3 \cos(\frac{\pi}{4} - 2\omega t)$

By applying the trigonometric identity $\cos(-\theta) = \cos(\theta)$, we can simplify the argument of the function:

$f(t) = 3 \cos(2\omega t - \frac{\pi}{4})$

[Per the properties of circular functions and trigonometric symmetry].

Step 2: Determining the Nature of Motion

A function represents Simple Harmonic Motion (SHM) if it can be expressed in the form $f(t) = A \cos(\omega' t + \phi)$ or $f(t) = A \sin(\omega' t + \phi)$, where $A$ is the amplitude, $\omega'$ is the angular frequency, and $\phi$ is the phase constant.

Comparing our given function $f(t) = 3 \cos(2\omega t - \frac{\pi}{4})$ with the standard form:

• Amplitude $A = 3$
• Angular frequency $\omega' = 2\omega$
• Phase constant $\phi = -\frac{\pi}{4}$

Since the function is a single-frequency sinusoidal function, it satisfies the requirements for Simple Harmonic Motion [Per the definition of linear SHM, which is the projection of uniform circular motion on a diameter].

Step 3: Calculating the Period of Motion

For a function of the form $\cos(\omega' t + \phi)$, the time period $T$ is given by the relation:

$T = \frac{2\pi}{\omega'}$

Substituting $\omega' = 2\omega$ into the formula:

$T = \frac{2\pi}{2\omega}$

$T = \frac{\pi}{\omega}$

[By the relation between angular frequency and temporal period in periodic motion].

Step 4: Visual Representation

Final Solution: The function represents Simple Harmonic Motion (SHM) with a time period $T = \frac{\pi}{\omega}$.

Step 1: System Definition and Initial Equilibrium

Consider a U-tube containing mercury of density $\rho$. Let $A$ be the uniform cross-sectional area of the tube. When the pressure difference is maintained, the mercury levels in the two arms are at different heights. Let the equilibrium position be $y = 0$, where the height of the mercury column in each arm is equal. When the suction pump is removed, the mercury column is displaced by a small vertical distance $y$ from the equilibrium position.

Step 2: Analysis of Restoring Force

When the mercury column is displaced by a distance $y$ in one arm, the level drops by $y$ in the other arm. Consequently, the total difference in the height of the mercury columns in the two arms becomes $h = y - (-y) = 2y$.

The weight of the excess mercury column, which acts as the restoring force $F$, is given by: $F = -(\text{Volume}) \times (\text{Density}) \times (g)$ Since the total length of the displaced column is $2y$, the volume is $V = A \times (2y)$. Thus, $F = -(A \cdot 2y) \rho g$ [Per the definition of weight of a liquid column].

Step 3: Equation of Motion

Let $L$ be the total length of the mercury column in the U-tube. The total mass $m$ of the mercury in the tube is: $m = \rho \cdot A \cdot L$ [Given mass is density times volume].

Applying Newton's Second Law of Motion ($F = ma$): $(\rho \cdot A \cdot L) \frac{d^2y}{dt^2} = -2Ay\rho g$

By simplifying the terms $A$ and $\rho$ from both sides, we obtain the differential equation: $\frac{d^2y}{dt^2} = -\left( \frac{2g}{L} \right)y$

Step 4: Comparison with Simple Harmonic Motion (SHM)

The standard differential equation for simple harmonic motion is: $\frac{d^2y}{dt^2} = -\omega^2 y$ [Where $\omega$ is the angular frequency].

Comparing the derived equation with the standard form, we identify the angular frequency: $\omega^2 = \frac{2g}{L} \implies \omega = \sqrt{\frac{2g}{L}}$

Since $\frac{d^2y}{dt^2} \propto -y$, the restoring force is directly proportional to the displacement and directed towards the equilibrium position. This satisfies the necessary and sufficient condition for Simple Harmonic Motion.

Step 5: Time Period of Oscillation

The time period $T$ of the oscillations of the mercury column is determined by: $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{2g}}$

Final Solution: The mercury column executes Simple Harmonic Motion because the restoring force $F = -(2A\rho g)y$ is proportional to the displacement $y$ and opposes the motion, conforming to the characteristic differential equation $\frac{d^2y}{dt^2} + \omega^2 y = 0$.

Step 1: Identifying Given Parameters and Physical Constants

Based on the provided motion parameters, we identify the following variables:

• Amplitude ($A$) = $5 \text{ cm} = 0.05 \text{ m}$
• Period ($T$) = $0.2 \text{ s}$
• Displacement ($x$) = $5 \text{ cm} = 0.05 \text{ m}$

Step 2: Calculating Angular Frequency ($\omega$)

The angular frequency is related to the period of oscillation by the relation $\omega = \frac{2\pi}{T}$ [Per the definition of SHM periodicity].

$\omega = \frac{2\pi}{0.2} = 10\pi \text{ rad/s}$

Step 3: Calculating Velocity ($v$)

The velocity of a particle in Simple Harmonic Motion is given by the expression $v = \omega \sqrt{A^2 - x^2}$ [By the conservation of energy in SHM].

Substituting the given values:

$v = 10\pi \sqrt{(0.05)^2 - (0.05)^2}$

$v = 10\pi \sqrt{0.0025 - 0.0025} = 0 \text{ m/s}$

Justification: At the extreme position ($x = A$), the body momentarily comes to rest before reversing its direction.

Step 4: Calculating Acceleration ($a$)

The acceleration of a particle in SHM is defined by $a = -\omega^2 x$ [Per the restoring force law $F = -kx$ and Newton's Second Law $F=ma$].

Substituting the values:

$a = -(10\pi)^2 \times (0.05)$

$a = -(100\pi^2) \times (0.05)$

$a = -5\pi^2 \text{ m/s}^2$

Using $\pi^2 \approx 9.87$:

$a \approx -5 \times 9.87 = -49.35 \text{ m/s}^2$

Visual Representation of SHM Extreme Position

Final Solution: At a displacement of 5 cm, the velocity is 0 m/s and the acceleration is $-5\pi^2 \text{ m/s}^2 \approx -49.35 \text{ m/s}^2$.

Step 1: Identifying the General Equation of Simple Harmonic Motion (SHM)

The displacement of a particle executing Simple Harmonic Motion is mathematically represented as:

$x(t) = A \cos(\omega t + \phi)$

[Where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant].

Step 2: Analyzing the Initial Conditions at $t = 0$

The problem states that at $t = 0$, the mass is at the maximum stretched position. Based on our coordinate system definition where the right direction is positive:

• At $t = 0$, the displacement $x = +A$.
• Substituting these values into the general equation:
• $A = A \cos(\omega(0) + \phi)$
• $1 = \cos(\phi)$
• $\phi = \arccos(1) = 0$ radians.

[By the definition of circular projection, the cosine function reaches its maximum value at an argument of zero].

Step 3: Constructing the Final Displacement Function

By substituting the phase constant $\phi = 0$ back into the general equation, we obtain the specific function for the given condition:

$x(t) = A \cos(\omega t + 0)$

$x(t) = A \cos(\omega t)$

Step 4: Visualizing the Motion

Step 5: Analytical Summary

Final Solution: The displacement of the mass as a function of time is $x(t) = A \cos(\omega t)$.

Theoretical Foundation of Simple Harmonic Motion (SHM)

According to the dynamical definition of Simple Harmonic Motion, a particle undergoes SHM if its acceleration $a$ is directly proportional to its displacement $x$ from a fixed equilibrium position and is always directed towards that equilibrium position. Mathematically, this is expressed as:

$a(x) = -\omega^2 x$

where $\omega$ is the angular frequency of the motion, and the negative sign indicates that the acceleration is always in the opposite direction of the displacement [Per Newton's Second Law and Hooke's Law].

Step 1: Analyzing the Given Equation

The provided relationship is:

$a = 0.7x$

In this expression, the constant of proportionality is positive ($+0.7$).

Step 2: Comparing with the SHM Criterion

We compare the given equation $a = 0.7x$ with the standard form $a = -\omega^2 x$.

• The standard form requires the acceleration to be directed towards the mean position (restoring force must be negative relative to displacement).
• In the given equation $a = 0.7x$, a positive displacement $x$ results in a positive acceleration $a$. This indicates that the force acts in the same direction as the displacement.
• This represents a repulsive force, which pushes the particle further away from the equilibrium position rather than restoring it [By definition of stable vs. unstable equilibrium].

Step 3: Visual Representation of the Relationship

Step 4: Conclusion

Since the acceleration $a$ does not oppose the displacement $x$ (i.e., there is no negative sign), the particle will move away from the origin with increasing acceleration if displaced, rather than oscillating about it. Therefore, the condition for SHM is not satisfied.

Final Solution: The relationship $a = 0.7x$ does not involve simple harmonic motion because the acceleration is not directed towards the equilibrium position (it lacks the necessary restoring force characteristic).

Step 1: Establishing the Coordinate System

In linear Simple Harmonic Motion (SHM), the equilibrium position $O$ is defined as the midpoint between the two extreme points $A$ and $B$. Let the total distance $AB = 10 \text{ cm}$. Therefore, the amplitude $A$ of the motion is:

$A = \frac{AB}{2} = 5 \text{ cm}$

Given the direction from $A \to B$ is positive ($+$), we define the positions as follows:

• Position $A$: $x = -5 \text{ cm}$
• Position $O$ (Midpoint): $x = 0 \text{ cm}$
• Position $B$: $x = +5 \text{ cm}$

Step 2: Analysis of the Particle's State

The particle is currently at the midpoint ($x = 0$) and is moving towards $A$. Since the direction $A \to B$ is defined as the positive direction, motion from $B \to A$ or towards $A$ is defined as the negative direction.

Step 3: Determination of Signs

Step 4: Theoretical Justification

The restoring force in SHM is given by $F = -kx$. At the equilibrium position ($x=0$), the displacement is zero, resulting in zero restoring force. By the definition of acceleration in SHM ($a = F/m$), the acceleration must also be zero. The velocity is non-zero (maximum) because the particle is passing through the equilibrium point, but its sign is dictated strictly by the displacement vector relative to the chosen positive axis. [Ref: Hooke's Law and the differential equation of SHM: $\frac{d^2x}{dt^2} + \omega^2 x = 0$].

Final Solution: At the mid-point of AB going towards A, the velocity is negative, the acceleration is zero, and the force is zero.

Step 1: Identifying Effective Acceleration

To determine the time period of a simple pendulum in a non-inertial frame (the accelerating car), we must first identify the net effective acceleration acting on the bob. The car is moving in a circular path of radius $R$ at a uniform speed $v$. This implies a centripetal acceleration ($a_c$) directed toward the center of the circular track:

$a_c = \frac{v^2}{R}$ [Per Newton's Second Law of Motion in a rotating frame]

In the frame of the car, the bob experiences a pseudo-force directed radially outward, equal in magnitude to $M \frac{v^2}{R}$. Simultaneously, the bob is subject to the gravitational force $Mg$ acting vertically downward.

Step 2: Vector Addition of Accelerations

The total effective acceleration ($g_{eff}$) is the vector sum of the gravitational acceleration ($\vec{g}$) and the pseudo-acceleration ($\vec{a}_p = -\vec{a}_c$):

$\vec{g}_{eff} = \vec{g} + \vec{a}_p$

Since the gravitational acceleration $\vec{g}$ (downward) and the pseudo-acceleration $\vec{a}_p$ (radially outward) are mutually perpendicular, the magnitude of $g_{eff}$ is derived using the Pythagorean theorem [By the Law of Vector Addition]:

$g_{eff} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}$

Step 3: Calculating the Time Period

The time period $T$ of a simple pendulum is governed by the formula $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$. Substituting our expression for $g_{eff}$ into the standard pendulum equation [Per Simple Harmonic Motion dynamics]:

$T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}}}$

Visual Representation of Forces

Step 4: Final Derivation

By simplifying the algebraic expression under the radical, we obtain the finalized period of oscillation for the pendulum in the moving car frame:

$T = 2\pi \left( l^2 \right)^{1/4} \cdot \left( g^2 + \frac{v^4}{R^2} \right)^{-1/4}$

Or, expressed in the standard radical form:

$T = 2\pi \sqrt{\frac{l}{(g^2 + v^4/R^2)^{1/2}}}$

Final Solution: $T = 2\pi \left[ \frac{l^2}{g^2 + (v^2/R)^2} \right]^{1/4}$