Simple Harmonic Motion

The potential energy $(U)$ of a particle executing simple harmonic motion $(S H M)$ is given by :

$ \mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2 $

Substituting the displacement equation $\mathrm{x}(\mathrm{t})=\mathrm{A} \sin \omega \mathrm{t}$ :

$ \mathrm{U}(\mathrm{t})=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2 \sin ^2 \omega \mathrm{t} $

The potential energy $U$ is maximum when the displacement $x$ is maximum $(x= \pm A)$.

This occurs when :

$ \sin ^2 \omega t=1 \Rightarrow \sin \omega t= \pm 1 $

This occurs for a particle starting from the mean position ( $t=0$ ) is at :

$ \omega \mathrm{t}=\frac{\pi}{2} $

Using the relationship between angular frequency $\omega$ and time period (T), $\omega=\frac{2 \pi}{\mathrm{~T}}$

$ \Rightarrow\left(\frac{2 \pi}{T}\right) t=\frac{\pi}{2} \Rightarrow t=\frac{T}{4} $

It is given that $\mathrm{t}=\frac{\mathrm{T}}{2 \beta}$,

$ \frac{T}{4}=\frac{T}{2 \beta} \Rightarrow \beta=2 $

Therefore, the value of $\beta$ is 2 . So, the correct answer is 2 .

For a particle in simple harmonic motion, the velocity is given by

$ v^2=\omega^2\left(A^2-x^2\right) $

We are given

$ v^2=50-x^2 $

Comparing this with the standard SHM form,

$ v^2=\omega^2(A^2-x^2) $

we get

$ \omega^2=1 $

So,

$ \omega=1\ \text{rad/s} $

Now, time period of SHM is

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

Therefore,

$ T=\frac{2\pi}{1}=2\pi\ \text{s} $

Given in the question that time period is

$ T=\frac{x}{7}\ \text{s} $

So,

$ \frac{x}{7}=2\pi $

Hence,

$ x=14\pi $

Using $\pi=\frac{22}{7}$,

$ x=14\times \frac{22}{7}=44 $

So, the required value is

$ \boxed{44} $

Let a uniform thin rod of mass $M$, length $L$, cross-sectional area $A$, and density $\rho$ be hinged at the bottom of the tank (origin $O$). The total mass of the rod can be expressed as:

$M = \rho A L$

The lower layer has a density $\rho_1 = 6\rho$ and extends from $y = 0$ to $y = \frac{L}{2}$

The upper layer has a density $\rho_2 = 2\rho$ and extends from $y = \frac{L}{2}$ to $y = L$.

When the rod is tilted by a small angle $\theta$ relative to the vertical axis, the forces exert a torque about the hinge $O$:

Gravitational force $F_g = Mg$ is acting downwards at the center of mass.

Buoyant force $(F_{B1})$ from the lower heavy liquid layer.

Buoyant Force $(F_{B2})$ from the upper lighter liquid layer.

The mass of the rod is uniformly distributed, so its center of mass lies exactly at its geometric center, which is at a distance of $\frac{L}{2}$ from the hinge $O$.

When tilted by an angle $\theta$, the restoring torque due to gravity tending to pull the rod back towards the vertical line is :

$\tau_g = -Mg\left(\frac{L}{2}\right)\sin\theta$

Using the small-angle approximation $\sin\theta \approx \theta$ :

$ \tau_g=-(\rho \mathrm{AL}) \mathrm{g} \frac{\mathrm{~L}}{2} \theta=-\frac{1}{2} \rho \mathrm{Ag} \mathrm{~L}^2 \theta $

The buoyant force on any submerged segment of a body equals the weight of the fluid displaced by that segment. Since the buoyant force acts upwards, it pushes the rod away from the vertical axis, creating an overturning torque that opposes gravity.

For lower segment $\mathrm{y}=0$ to $\mathrm{y}=\frac{\mathrm{L}}{2}$ :

The length of this segment inside the lower liquid is $\mathrm{l}_1=\frac{\mathrm{L}}{2}$.

The mass of the displaced liquid is $\mathrm{M}_{\mathrm{fluid} 1}=\rho_1 \mathrm{Al}_1=(6 \rho) \mathrm{A}\left(\frac{\mathrm{L}}{2}\right)=3 \rho \mathrm{AL}$.

So, the buoyant force is :

$ \mathrm{F}_{\mathrm{B} 1}=3 \rho \mathrm{ALg} $

Since the segment is uniform, this force acts at its midpoint, which is at a distance of $\mathrm{y}_1=\mathrm{L} 4$ from the hinge O.

The torque due to $\mathrm{F}_{\mathrm{B} 1}$ about the hinge is :

$ \tau_{\mathrm{B} 1}=+\mathrm{F}_{\mathrm{B} 1} \cdot \mathrm{y}_1 \sin \theta \approx+(3 \rho \mathrm{ALg})\left(\frac{\mathrm{L}}{4}\right) \theta=+\frac{3}{4} \rho \mathrm{Ag} \mathrm{~L}^2 \theta $

For the upper segment $y=\frac{L}{2}$ to $y=L$ :

The length of this segment inside the upper liquid is $\mathrm{l}_2=\frac{\mathrm{L}}{2}$.

The mass of the displaced liquid is $\mathrm{M}_{\text {fluid2 }}=\rho_2 \mathrm{Al}_2=(2 \rho) \mathrm{A}\left(\frac{\mathrm{L}}{2}\right)=\rho \mathrm{AL}$.

So, the buoyant force is :

$ \mathrm{F}_{\mathrm{B} 2}=\rho \mathrm{ALg} $

This force acts at the midpoint of the upper segment. The distance of this midpoint from the hinge O is :

$ \mathrm{y}_2=\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{4}=\frac{3 \mathrm{~L}}{4} $

The torque due to $\mathrm{F}_{\mathrm{B} 2}$ about the hinge is :

$\tau_{B2} = +F_{B2} \cdot y_2 \sin \theta \approx +(\rho A L g)\left(\frac{3L}{4}\right)\theta = +\frac{3}{4}\rho A g L^2\theta$

The net restoring torque $\tau_{\text{net}}$ acting on the rod about the hinge O is

$\tau_{\text{net}} = \tau_g + \tau_{B1} + \tau_{B2}$

$\tau_{\text{net}} = -\frac{1}{2}\rho A g L^2\theta + \frac{3}{4}\rho A g L^2\theta + \frac{3}{4}\rho A g L^2\theta$

$\tau_{\text{net}} = \left(-\frac{1}{2} + \frac{3}{4} + \frac{3}{4}\right)\rho A g L^2\theta$

$\tau_{\text{net}} = +1\cdot \rho A g L^2\theta$

The positive sign indicates that the net torque is in the direction opposite to the angular displacement $\theta$

$\tau_{\text{restoring}} = -\rho A g L^2\theta$

The moment of inertia $I$ of a uniform thin rod of mass $M$ and length $L$ rotating about an axis passing through one of its ends is derived as :

$I = \frac{1}{3}ML^2$

Substituting $M = \rho A L$:

$I = \frac{1}{3}(\rho A L)L^2 = \frac{1}{3}\rho A L^3$

The equation of rotational simple harmonic motion is given by:

$I\alpha = \tau_{\text{restoring}} \Rightarrow I\frac{d^2\theta}{dt^2} = -\rho A g L^2\theta$

$\left(\frac{1}{3}\rho A L^3\right)\frac{d^2\theta}{dt^2} = -\rho A g L^2\theta$

$\frac{1}{3}L\frac{d^2\theta}{dt^2} = -g\theta \Rightarrow \frac{d^2\theta}{dt^2} + \left(\frac{3g}{L}\right)\theta = 0$

Comparing with standard SHM equation $\frac{d^2\theta}{dt^2} + \omega^2\theta = 0$, the angular frequency $\omega$ is :

$\omega = \sqrt{\frac{3g}{L}}$

So, the time period $T$ of the small oscillations is :

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{3g}} = \frac{2\pi}{\sqrt{3}}\sqrt{\frac{L}{g}}$

$\Rightarrow $ $ \mathrm{T}=\frac{2 \pi}{\mathrm{n}} \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}=\frac{2 \pi}{\sqrt{3}} \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}} \Rightarrow \mathrm{n}=\sqrt{3} \approx 1.73 $

Therefore, the correct answer is 1.73

Here, we will use the formula for apparent weight in elevator as the elevator has some acceleration.

The normal force (apparent weight) adjusts for the elevations acceleration

relative to gravity.

Apparent weight: $w=m(g+a)$, where $a$ is the elevator's acceleration.

Here, elevator's position, $y$ is given so we can find elevator's acceleration using

$ a=\frac{d^2 y}{d t^2} $

Given, $m=50 \mathrm{~kg}$

$ \begin{aligned} & \text { and } y=8\left[1+\sin \left(\frac{2 \pi t}{T}\right)\right] \\ & T=40 \pi \mathrm{~s} . \\ & \Rightarrow y=8(1+\sin (\omega t)) \quad\left(as, \omega=\frac{2 \pi}{T}\right) \\ & \omega=\frac{2 \pi}{40 \pi}=\frac{1}{20} \\ & a=\frac{d^2 y}{d t^2}=-8 \omega^2 \sin (\omega t) \\ & \left|a_{\text {max }}\right|=8 \omega^2=8\left(\frac{1}{20}\right)^2=8 \times \frac{1}{400} \\ & \left|a_{\text {max }}\right|=\frac{1}{50} \end{aligned} $

$W_{\text {max }}=m\left(g+a_{\text {max }}\right)$ (when lift goes up)

and. $W_{\text {min }}=m\left(g-a_{\text {max }}\right)$ (when lift goes down)

so

$ \begin{aligned} \Delta w & =w_{\text {max }}-w_{\text {min }}=2 m a_{\text {max }} \\ & =2 \times 50 \times \frac{1}{50} \Rightarrow \Delta w=2 \mathrm{~N} \end{aligned} $

To find the value of $x$ that represents the time period of oscillation in this simple harmonic motion (SHM) scenario, we first recall the general formula for the time period ($T$) of a mass-spring system undergoing SHM, which is given by:

$T = 2\pi \sqrt{\frac{m}{k}}$

Here,

$m$ is the mass of the particle, which is $0.50 \, \mathrm{kg}$ in this case,

$k$ is the force constant of the spring or the spring constant, which is given as $50 \, \mathrm{Nm^{-1}}$,

and $T$ represents the time period of oscillation.

Given in the problem, $T = \frac{x}{35} \, \mathrm{s}$ and we are provided with the approximation $\pi = \frac{22}{7}$.

Substituting the given values into the formula for $T$:

$\frac{x}{35} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.50}{50}}$

To simplify this, we first calculate the square root:

$\sqrt{\frac{0.50}{50}} = \sqrt{\frac{1}{100}} = \frac{1}{10}$

Substituting back, we get:

$\frac{x}{35} = 2 \times \frac{22}{7} \times \frac{1}{10}$

Multiplying the terms on the right side:

$\frac{x}{35} = \frac{44}{70}$

$\frac{x}{35} = \frac{22}{35}$

Multiplying both sides by $35$ to solve for $x$:

$x = 22$

Therefore, the value of $x$ that represents the time period of oscillation is $22$ seconds.

Let's begin by understanding the equations related to simple harmonic motion (SHM). For a particle executing SHM, the position $x$, velocity $v$, and acceleration $a$ are given by the following equations:

1. Position: $x = A \cos(\omega t + \phi)$

2. Velocity: $v = -A \omega \sin(\omega t + \phi)$

3. Acceleration: $a = -A \omega^2 \cos(\omega t + \phi)$

Here, $A$ is the amplitude of the motion, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Given the magnitudes at a certain instant:

$x = 4 \, \mathrm{m}$

$v = 2 \, \mathrm{ms}^{-1}$

$a = 16 \, \mathrm{ms}^{-2}$

Using the acceleration equation:

$a = -A \omega^2 \cos(\omega t + \phi)$

Since we’re given the magnitude of the acceleration, we remove the negative sign:

$16 = A \omega^2 \cos(\omega t + \phi)$

Using the position equation:

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

We already know $x = 4 \, \mathrm{m}$, so:

$4 = A \cos(\omega t + \phi)$

From these two equations, we know:

$A \omega^2 \cos(\omega t + \phi) = 16$

$A \cos(\omega t + \phi) = 4$

Therefore:

$A \omega^2 \cdot 4/A = 16$

$4 \omega^2 = 16$

$\omega^2 = 4$

$\omega = 2 \, \mathrm{rad/s}$

Next, using the velocity equation:

$v = -A \omega \sin(\omega t + \phi)$

Again, we consider the magnitude:

$2 = A \cdot 2 \sin(\omega t + \phi)$

$2 = 2A \sin(\omega t + \phi)$

$\sin(\omega t + \phi) = \dfrac{1}{A}$

We know from the position equation that:

$\cos(\omega t + \phi) = \dfrac{4}{A}$

Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we get:

$\left(\dfrac{1}{A}\right)^2 + \left(\dfrac{4}{A}\right)^2 = 1$

$\dfrac{1}{A^2} + \dfrac{16}{A^2} = 1$

$\dfrac{17}{A^2} = 1$

$A^2 = 17$

$A = \sqrt{17} \, \mathrm{m}$

Therefore, the amplitude of the motion is $\sqrt{17} \, \mathrm{m}$, meaning $x$ is 17.

To solve for the amplitude of oscillation, we start by using the properties of simple harmonic motion (SHM). In SHM, the total energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE).

Given:

• Mass $m = 0.2 \ \mathrm{kg}$
• Frequency $f = \left(\frac{25}{\pi}\right) \ \mathrm{Hz}$
• Position $x = 0.04 \ \mathrm{m}$
• KE at $x = 0.04 \ \mathrm{m}$ is $0.5 \ \mathrm{J}$
• PE at $x = 0.04 \ \mathrm{m}$ is $0.4 \ \mathrm{J}$

The total mechanical energy (E) of the SHM system can be found by summing the given kinetic and potential energies:

$ E = KE + PE = 0.5 \ \mathrm{J} + 0.4 \ \mathrm{J} = 0.9 \ \mathrm{J} $

For simple harmonic motion, the total energy (E) is also related to the amplitude (A) by the following formula:

$ E = \frac{1}{2} k A^2 $

where $k$ is the spring constant. First, we need to find the angular frequency $\omega$:

$ \omega = 2 \pi f = 2 \pi \left(\frac{25}{\pi}\right) \ \mathrm{Hz} = 50 \ \mathrm{rad/s} $

The spring constant $k$ can be calculated using the relationship between $m$, $\omega$, and $k$:

$ \omega = \sqrt{\frac{k}{m}} \Rightarrow k = m \omega^2 = 0.2 \times (50)^2 = 500 \ \mathrm{N/m} $

Now, substituting $k$ back into the energy equation, we solve for the amplitude $A$:

$ 0.9 = \frac{1}{2} \times 500 \times A^2 \Rightarrow A^2 = \frac{0.9 \times 2}{500} \Rightarrow A^2 = \frac{1.8}{500} \Rightarrow A^2 = 0.0036 \Rightarrow A = \sqrt{0.0036} = 0.06 \ \mathrm{m} $

Converting $A$ from meters to centimeters:

$ A = 0.06 \ \mathrm{m} \times 100 = 6 \ \mathrm{cm} $

Thus, the amplitude of oscillation is $6 \ \mathrm{cm}$.

For a particle performing simple harmonic motion (SHM), the maximum velocity $v_{max}$ can be calculated using the formula:

$v_{max} = A\omega$

where $A$ is the amplitude of the motion and $\omega$ is the angular frequency. The angular frequency $\omega$ is related to the time period $T$ by the formula:

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

Given:

• Amplitude, $A = 0.06 \, \mathrm{m}$


• Time period, $T = 3.14 \, \mathrm{s}$

First, we find the angular frequency:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14}$

Substituting $\omega$ and $A$ in the formula for $v_{max}$:

$v_{max} = A\omega = 0.06 \times \frac{2\pi}{3.14}$

$v_{max} = 0.06 \times \frac{2 \times 3.14}{3.14}$

$v_{max} = 0.06 \times 2$

$v_{max} = 0.12 \, \mathrm{m/s}$

To convert meters per second to centimeters per second, we use the conversion factor $1 \, \mathrm{m/s} = 100 \, \mathrm{cm/s}$. Therefore,

$v_{max} = 0.12 \, \mathrm{m/s} \times 100 \, \mathrm{cm/m} = 12 \, \mathrm{cm/s}$

Thus, the maximum velocity of the particle is $12 \, \mathrm{cm/s}$.

The displacement of a particle executing Simple Harmonic Motion (SHM) can be expressed as:

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

Where:

$A$ is the amplitude of the SHM,

$\omega$ is the angular frequency,

$t$ is the time,

$\phi$ is the phase constant (phase angle at $t = 0$).

In the given equation, $x = 10 \sin(\omega t + \frac{\pi}{3})$ m, the amplitude $A = 10$ m and the phase constant $\phi = \frac{\pi}{3}$. The time period $T = 3.14$ s is given, from which we can find the angular frequency $\omega$ using the relationship:

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

Substituting the given $T = 3.14$ s:

$\omega = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$

To find the velocity of the particle, we differentiate the displacement $x$ with respect to time $t$. The derivative of the displacement gives the velocity:

$v = \frac{dx}{dt}$

So, for $x = 10 \sin(\omega t + \frac{\pi}{3})$:

$v = \frac{d}{dt}[10 \sin(\omega t + \frac{\pi}{3})]$

Applying differentiation, we get:

$v = 10\omega \cos(\omega t + \frac{\pi}{3})$

Plug in the value of $\omega = 2$ rad/s and evaluate it at $t = 0$ to find the initial velocity:

$v = 10 \cdot 2 \cos(2 \cdot 0 + \frac{\pi}{3})$

$v = 20 \cos(\frac{\pi}{3})$

$\cos(\frac{\pi}{3}) = \frac{1}{2}$, therefore:

$v = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$

Thus, the velocity of the particle at $t = 0$ is $10$ m/s.

Let's start by considering the formula for the frequency of a mass on a spring (a simple harmonic oscillator):

$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $

Where:

• $ f $ is the frequency of oscillation
• $ k $ is the spring constant
• $ m $ is the mass suspended from the spring

When the mass $ m $ is suspended, the frequency $ f_1 $ is:

$ f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $

When the mass $ 9m $ is suspended, the frequency $ f_2 $ is:

$ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}} $

We can simplify the square root by taking the 9 inside the root as $3^2$, which gives:

$ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{(3^2)m}} $

$ f_2 = \frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}} $

The ratio of $ \frac{f_1}{f_2} $ is therefore:

$ \frac{f_1}{f_2} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{m}}}{\frac{1}{2\pi} \frac{1}{3} \sqrt{\frac{k}{m}}} $

$ \frac{f_1}{f_2} = \frac{1}{\frac{1}{3}} $

$ \frac{f_1}{f_2} = 3 $

So the value of $ \frac{f_1}{f_2} $ is $3$.

$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$

Angular frequency of oscillation $(\omega)=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}$

$(\omega)=\sqrt{\frac{5 \mathrm{k}}{3 \mathrm{~m}}}$

Period of oscillation $(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \mathrm{~m}}{5 \mathrm{k}}}$

$=\pi \sqrt{\frac{12 \mathrm{~m}}{5 \mathrm{k}}}$

To find the new amplitude of the motion when the speed is increased to three times at a given displacement, we use the concepts of simple harmonic motion (SHM) and its formulas.

In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position can be given by the formula:

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

where:

• $\omega$ is the angular frequency of the motion,
• $A$ is the amplitude, and
• $x$ is the displacement at that instance.

Given:

• Displacement at the instance, $x = \frac{2A}{3}$,
• Initial velocity is increased to three times at this displacement.

Thus, let's find the initial velocity $v$ at $x = \frac{2A}{3}$:

$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$

With the velocity increased to three times, the new velocity $v'$ becomes:

$v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$

For the new amplitude $A'$, the velocity $v'$ at the same displacement $x$ is:

$v' = \omega \sqrt{{A'}^2 - \left(\frac{2A}{3}\right)^2}$

Setting the expressions for $v'$ equal gives:

$\sqrt{5}A\omega = \omega \sqrt{{A'}^2 - \frac{4A^2}{9}}$

$\sqrt{5}A = \sqrt{{A'}^2 - \frac{4A^2}{9}}$

Solving for $A'$ gives:

${A'}^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$

$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$

Therefore, the new amplitude of the motion is $\frac{7A}{3}$, which means the value of $n$ is 7.

From phasor diagram particle has to move from $\mathrm{P}$ to $\mathrm{Q}$ in a circle of radius equal to amplitude of SHM.

$\begin{aligned} & \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \\ & \phi=\frac{\pi}{6} \end{aligned}$

Now, $\frac{\pi}{6}=\omega \mathrm{t}$

$\begin{aligned} & \frac{\pi}{6}=\frac{2 \pi}{T} t \\ & \frac{\pi}{6}=\frac{2 \pi}{6 \pi} t \end{aligned}$

$\mathrm{t}=\frac{\pi}{2}$

So, $x=2$

$\begin{aligned} & \text { Let total energy }=\mathrm{E}=\frac{1}{2} \mathrm{KA}^2 \\ & \mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^2=\frac{\mathrm{KA}^2}{2 \times 9}=\frac{\mathrm{E}}{9} \\ & \mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9} \\ & \text { Ratio } \frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8} \\ & \mathrm{x}=9 \end{aligned}$

$\begin{aligned} & \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\ & \quad \omega=\frac{5}{2} \\ & \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\ & \mathrm{x}=\sqrt{12} \mathrm{~cm} \end{aligned}$

At T t$_0$ = 0

Before collision

After collision

$\begin{aligned} \mathrm{v}_{\mathrm{CM}} & =\frac{\mathrm{m} \cdot \frac{\mathrm{a} \omega}{2}+\mathrm{m} \cdot \mathrm{a} \omega}{\mathrm{m}+\mathrm{m}} \\ \mathrm{v}_{\mathrm{CM}} & =\frac{3 \mathrm{a} \omega}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =\frac{3}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =0.75 \end{aligned}$

$\mathrm{t}_0=\frac{\pi}{2 \omega}=\frac{\mathrm{T}}{4}$

Particles are at extreme position

After collision

in C-frame

using WET,

$\begin{aligned} & \mathrm{W}_{\text {spring }}=\Delta \mathrm{K} \\ & \frac{1}{2} \mathrm{k}(2 \mathrm{~b})^2-\frac{1}{2} \mathrm{k}(2 \mathrm{a})^2=2 \times \frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{a} \omega}{4}\right)^2 \quad(\mathrm{k}=\text { spring constant }) \\ & 4 \mathrm{~kb}^2-4 \mathrm{ka}^2=2 \times \mathrm{m} \times \frac{\mathrm{a}^2}{16} \times \frac{2 \mathrm{k}}{\mathrm{m}} \\ & 4 \mathrm{~b}^2=\frac{17}{4} \mathrm{a}^2 \\ & \frac{4 \mathrm{~b}^2}{\mathrm{a}^2}=4.25 \end{aligned}$

To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.

First, let's find the angular frequency $\omega$:

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

where $T = 3.14\,\mathrm{s}$ is the time period.

$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$

Now, let's find the maximum velocity $v_{max}$ of the block:

$v_{max} = \omega A$

where $A = 1\,\mathrm{m}$ is the amplitude.

$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$

Next, we can find the spring constant $k$ using the mass of the block $m = 5\,\mathrm{kg}$ and the angular frequency $\omega$:

$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$

$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$

Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:

$F_{max} = kA$

$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$

The maximum force exerted by the spring on the block is $20\,\mathrm{N}$.

Given the amplitude $A$ and the length $l$ of the pendulum, we can find the maximum angular displacement $\theta_0$:

$ \sin \theta_0 = \frac{A}{l} = \frac{10}{100} = \frac{1}{10} $

By conservation of energy, the following equation holds:

$ \frac{1}{2} m v^2 = m g l(1 - \cos \theta) $

The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:

$ \begin{aligned} & T - mg = \frac{m v^2}{l} \ & \Rightarrow T = mg + \frac{m v^2}{l} \end{aligned} $

Substituting the conservation of energy equation, we get:

$ \begin{aligned} & T = mg + 2 m g(1 - \cos \theta) \\\\ & = mg\left[1 + 2\left(1 - \sqrt{1 - \sin^2 \theta}\right)\right] \\\\ & = mg\left[3 - 2 \sqrt{1 - \frac{1}{100}}\right] \\\\ & = \frac{250}{1000} \times 10\left[3 - 2\left(1 - \frac{1}{200}\right)\right] = \frac{101}{40} \\\\ & \therefore x = 101 \end{aligned} $

So, the value of $x$ is 101.

$4{v^2} = 50 - {x^2}$

or $v = {1 \over 2}\sqrt {50 - {x^2}} $

Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}} $

$ \Rightarrow \omega = {1 \over 2}$

& $A = \sqrt {50} $

so ${{2\pi } \over T} = {1 \over 2}$

$ \Rightarrow T = 4\pi \sec $

$ = 4 \times {{22} \over 7}\sec $

$T = {{88} \over 7}\sec $

so $x = 88$

$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$

So, ${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$

This value will be maximum when $\sin 2\omega t = 1$

or $2\omega t = {\pi \over 2}$

$2 \times {{2\pi } \over T}t = {\pi \over 2}$

$ \Rightarrow t = {T \over 8}$

So $\beta = 8$

$F = - 25x$

$.250{{{d^2}x} \over {d{t^2}}} = - 25x$

${{{d^2}x} \over {d{t^2}}} = - 100x$

$ \Rightarrow \omega = 10$ rad/sec

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

$10\,A = 4$

$ \Rightarrow A = 0.4$ m

$ = 40$ cm

$U = 4(1 - \cos 4x)$

$ \Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$

$ = - 16\sin 4x$

as small x

$F = - 16(4x) = - 64x \equiv - kx$

$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$

$ \Rightarrow K = 2$

$(0.9){A_1}\sqrt {{K \over {0.9}}} = (0.9 + 0.124){A_2}\sqrt {{K \over {0.9 + 0.124}}} $

${{{A_1}} \over {{A_2}}} = \sqrt {{{0.9 + 0.124} \over {0.9}}} $

$ = \sqrt {{{1.024} \over {0.9}}} $

$ = {\alpha \over {\alpha - 1}}$

$\alpha = 16$

For case (a),

${K_{eq}} = {{2K} \over 3}$

For case (b),

${K_{eq}} = 3K$

$\because$ $T = 2\pi \sqrt {{m \over K}} $

$\therefore$ ${{{T_a}} \over {{T_b}}} = \sqrt {{{{K_b}} \over {{K_a}}}} $

${3 \over {{T_b}}} = \sqrt {{{3K \times 3} \over {2K}}} = {3 \over {\sqrt 2 }}$

${T_b} = \sqrt 2 $

$x = 2$

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

$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $

$ \Rightarrow 9 \times 75 = {(A')^2} - 25$

$ \Rightarrow A' = \sqrt {28 \times 25} $ cm

$ \Rightarrow x = 700$

${1 \over 2}m{v^2} = mgl(1 - \cos \theta )$

$ \Rightarrow v = \sqrt {2gl(1 - \cos \theta )} $

$ = \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}} $

$ = 5$ m/s

A = 8 cm

T = 6 s

$A\cos \left( {{{2\pi t} \over T}} \right) = {A \over 2}$

$ \Rightarrow {{2\pi t} \over T} = {\pi \over 3}$

or $t = {T \over 6} = 1\,s$

When mass m is pulled by a force F, the wire elongation x, length l, cross-sectional area A, and Young's modulus of wire material Y are related by $Y = {{F/A} \over {x/l}}$ i.e.,

$F = (Y\,A/l)x$.

The restoring force by the wire is equal but opposite to F i.e., F_r = $-$F. Apply Newton's second law to get

$m{d^2}x/d{t^2} = - (YA/l)x = - {\omega ^2}x$.

This equation represents SHM with an angular frequency $\omega = \sqrt {YA/(lm)} $. Substitute the values to get $Y = {\omega ^2}lm/A = 4 \times {10^9}$ N/m².

To find the amplitude of oscillation, let's start by considering the given data and the formulae related to simple harmonic motion (SHM).

The total mechanical energy of a system in SHM is constant and is the sum of kinetic energy (KE) and potential energy (PE) at any point in its motion. We are given:

Kinetic Energy, $ KE = 0.5 \, \text{J} $

Potential Energy, $ PE = 0.4 \, \text{J} $

Therefore, the total energy (E) is:

$ E = KE + PE = 0.5 + 0.4 = 0.9 \, \text{J} $

In SHM, the total mechanical energy can also be expressed as:

$ E = \frac{1}{2} k A^2 $

where $ k $ is the spring constant and $ A $ is the amplitude.

We can also express kinetic and potential energy in SHM in terms of amplitude $ A $, position $ x $, and angular frequency $ \omega $:

Kinetic Energy: $ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) $

Potential Energy: $ PE = \frac{1}{2} m \omega^2 x^2 $

Given mass $ m = 0.2 \, \text{kg} $, position $ x = 0.04 \, \text{m} $.

The angular frequency $ \omega $ is related to the frequency $ f $ by:

$ \omega = 2\pi f $

Given frequency:

$ f = \frac{25}{\pi} \, \text{Hz} $

Thus,

$ \omega = 2\pi \left(\frac{25}{\pi}\right) = 50 \, \text{rad/s} $

Using the expression for the total energy in SHM:

$ \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 $

Since the total energy $ E = 0.9 \, \text{J} $, it follows:

$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot (50)^2 \cdot A^2 $

Solving for $ A^2 $:

$ 0.9 = \frac{1}{2} \cdot 0.2 \cdot 2500 \cdot A^2 $

$ 0.9 = \frac{1}{2} \cdot 500 \cdot A^2 $

$ 0.9 = 250 \cdot A^2 $

$ A^2 = \frac{0.9}{250} $

$ A^2 = 0.0036 $

$ A = \sqrt{0.0036} $

$ A = 0.06 \, \text{m} $

The amplitude of the oscillations is 0.06 m.