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.

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: 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: 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 Given Variables

To determine the spring constant $k$, we utilize the equilibrium condition of the automobile's mass on the suspension system.

• Total mass of the automobile, $M = 3000 \, \text{kg}$
• Sag (displacement), $x = 15 \, \text{cm} = 0.15 \, \text{m}$
• Acceleration due to gravity, $g \approx 9.8 \, \text{m/s}^2$

Step 2: Theoretical Foundation for Spring Constant

At static equilibrium, the restoring force of the spring system must balance the gravitational force acting on the automobile [Per Hooke's Law and Newton's Second Law].

$F_{restoring} = F_{gravity}$

$kx = Mg$

Step 3: Algebraic Manipulation and Calculation

Rearranging the equation to solve for the spring constant $k$:

$k = \frac{Mg}{x}$

Substitute the given numerical values:

$k = \frac{3000 \, \text{kg} \times 9.8 \, \text{m/s}^2}{0.15 \, \text{m}}$

$k = \frac{29400 \, \text{N}}{0.15 \, \text{m}}$

$k = 1.96 \times 10^5 \, \text{N/m}$

Visual Representation of Static Sag

Step 4: Theoretical Justification

The spring constant represents the stiffness of the suspension system. Since the four wheels share the load, this calculated $k$ represents the effective spring constant of the entire suspension system [Per the principle of equivalent spring constants in parallel].

Final Solution: The estimated value of the spring constant $k$ is $1.96 \times 10^5 \, \text{N/m}$.

Given Variables & Initial Setup

The system describes Simple Harmonic Motion (SHM) defined by the following parameters:

• Amplitude ($A$) = $5\text{ cm} = 0.05\text{ m}$
• Period ($T$) = $0.2\text{ s}$
• Displacement ($x$) = $3\text{ cm} = 0.03\text{ m}$

First, calculate the angular frequency ($\omega$) using the relationship between period and frequency [Per the definition of SHM]:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi\text{ rad/s} \approx 31.42\text{ rad/s}$

Step 1: Calculating Velocity ($v$)

The velocity of a particle in SHM as a function of displacement $x$ is given by [Derived from the conservation of energy in SHM]:

$v = \pm \omega \sqrt{A^2 - x^2}$

Substituting the given values:

$v = \pm (10\pi) \sqrt{(5)^2 - (3)^2}\text{ cm/s}$

$v = \pm 10\pi \sqrt{25 - 9} = \pm 10\pi \sqrt{16} = \pm 10\pi(4)$

$v = \pm 40\pi\text{ cm/s} \approx \pm 125.66\text{ cm/s}$

$v = \pm 1.26\text{ m/s}$

Step 2: Calculating Acceleration ($a$)

The acceleration of a particle in SHM is given by the differential equation of motion [Per Newton's Second Law and Hooke's Law]:

$a = -\omega^2 x$

Substituting the known values:

$a = -(10\pi)^2 \times (3\text{ cm})$

$a = -100\pi^2 \times 3$

$a = -300\pi^2\text{ cm/s}^2$

Using $\pi^2 \approx 9.87$:

$a \approx -300 \times 9.87 = -2961\text{ cm/s}^2 = -29.61\text{ m/s}^2$

Visual Representation of SHM Vectors

Summary Table

Final Solution: The velocity of the body is approximately $\pm 1.26\text{ m/s}$ and the acceleration is $-29.61\text{ m/s}^2$.

Step 1: Given Variables & Initial Setup

To analyze the Simple Harmonic Motion (SHM) of the body, we extract the primary parameters provided:

• Amplitude ($A$): $5\text{ cm} = 0.05\text{ m}$
• Period ($T$): $0.2\text{ s}$
• Displacement ($x$): $0\text{ cm} = 0\text{ m}$

We calculate the angular frequency ($\omega$) using the relationship between period and frequency [Per the definition of SHM]:

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

Step 2: Determining Velocity at Displacement $x=0$

The velocity ($v$) of a body in SHM is defined by the function [Derived from the differentiation of $x(t) = A \sin(\omega t + \phi)$]:

$v = \omega \sqrt{A^2 - x^2}$

Substituting $x = 0$ into the equation:

$v = (10\pi) \sqrt{(0.05)^2 - 0^2}$

$v = (10\pi) \times 0.05$

$v = 0.5\pi\text{ m/s} \approx 1.57\text{ m/s}$

Step 3: Determining Acceleration at Displacement $x=0$

The acceleration ($a$) of a body in SHM is defined by the restoring force relationship [Per Newton’s Second Law and Hooke’s Law]:

$a = -\omega^2 x$

Substituting $x = 0$ into the equation:

$a = -(10\pi)^2 \times 0$

$a = 0\text{ m/s}^2$

Visual Representation of SHM Parameters

Summary Analysis

At the equilibrium position ($x=0$), the body possesses maximum potential energy transformation into kinetic energy, resulting in maximum velocity. Conversely, since the restoring force $F = -kx$ is directly proportional to displacement, the acceleration vanishes at the equilibrium point.

Final Solution: The velocity is $1.57\text{ m/s}$ and the acceleration is $0\text{ m/s}^2$.

Step 1: Given Variables & System Parameters

For a single wheel system, we define the following parameters based on the problem statement:

• Mass per wheel, $m = 750\text{ kg}$
• Static sag, $x_0 = 0.15\text{ m}$
• Amplitude decay ratio: $A(T) = 0.5 A(0)$, where $T$ is the period of oscillation.

Step 2: Determining the Spring Constant ($k$)

The spring constant $k$ is derived from the equilibrium condition where the downward force of gravity equals the restoring spring force [Per Hooke's Law: $F = kx$]:

$mg = k x_0$

$k = \frac{mg}{x_0} = \frac{750 \times 9.8}{0.15}$

$k = \frac{7350}{0.15} = 49000\text{ N/m}$

Step 3: Analyzing Damped Harmonic Motion

For a damped oscillator, the displacement is given by $x(t) = A_0 e^{-(b/2m)t} \cos(\omega' t + \phi)$. The amplitude decays as $A(t) = A_0 e^{-(b/2m)t}$. [Per the theory of Damped Oscillations].

Given that after one period ($t = T$), the amplitude is $50\%$ of the initial amplitude:

$0.5 A_0 = A_0 e^{-(b/2m)T}$

$\ln(0.5) = -\frac{bT}{2m} \implies \frac{bT}{2m} = \ln(2) \approx 0.693$

Step 4: Calculating Period ($T$) and Damping Constant ($b$)

Assuming the damping is light ($\omega' \approx \omega_0 = \sqrt{k/m}$):

$\omega_0 = \sqrt{\frac{49000}{750}} \approx \sqrt{65.33} \approx 8.08\text{ rad/s}$

$T = \frac{2\pi}{\omega_0} = \frac{2 \times 3.1416}{8.08} \approx 0.777\text{ s}$

Now, solve for $b$ using the relationship derived in Step 3:

$b = \frac{2m \ln(2)}{T}$

$b = \frac{2 \times 750 \times 0.693}{0.777}$

$b = \frac{1039.5}{0.777} \approx 1337.8\text{ kg/s}$

Step 5: Visual Representation of Damped Decay

Final Solution: The estimated value of the damping constant $b$ is approximately $1338\text{ kg/s}$.

Step 1: Establishing the General Equation of Motion

The displacement of a simple harmonic oscillator is given by the general solution: $x(t) = A \cos(\omega t + \phi)$ where $A$ is the amplitude and $\phi$ is the initial phase constant. [Per the differential equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$]

Step 2: Defining Initial Conditions

At time $t = 0$, the initial position is $x(0) = x_0$ and the initial velocity is $v(0) = -v_0$ (negative as it is directed towards the center).
Using the displacement equation at $t=0$: $x_0 = A \cos(\phi)$ --- (Equation 1)

Differentiating $x(t)$ with respect to time to obtain the velocity equation: $v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)$

Applying the initial velocity condition at $t=0$: $-v_0 = -A \omega \sin(\phi)$ $v_0 = A \omega \sin(\phi)$ $\frac{v_0}{\omega} = A \sin(\phi)$ --- (Equation 2)

Step 3: Calculating the Amplitude

To isolate $A$, square Equation 1 and Equation 2 and invoke the trigonometric identity $\sin^2\phi + \cos^2\phi = 1$: $x_0^2 = A^2 \cos^2\phi$ $(\frac{v_0}{\omega})^2 = A^2 \sin^2\phi$

Summing the equations: $x_0^2 + \frac{v_0^2}{\omega^2} = A^2 (\cos^2\phi + \sin^2\phi)$ $x_0^2 + \frac{v_0^2}{\omega^2} = A^2$

Solving for $A$: $A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}}$

Step 4: Geometric Representation of Phase Space

Analytical Data Summary

Final Solution: $A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}}$