Simple Harmonic Motion

For a two-body system, the angular frequency ( $\omega$ ) is given by :

$ \omega=\sqrt{\frac{\mathrm{k}}{\mu}} $

Where :

• k is the spring constant.

• $\mu$ is the reduced mass of the system.

The reduced mass is calculated using the formula :

$ \mu=\frac{m_1 \cdot m_2}{m_1+m_2} $

Given values of masses are :

• $\mathrm{m}_1=1 \mathrm{~kg}$

• $\mathrm{m}_2=200 \mathrm{~g}=0.2 \mathrm{~kg}$

$ \mu=\frac{1 \times 0.2}{1+0.2}=\frac{0.2}{1.2}=\frac{1}{6} \mathrm{~kg} $

So the angular frequency is,

$ \omega=\sqrt{\frac{150}{1 / 6}} $

$\Rightarrow $ $ \omega=\sqrt{900}=30 \mathrm{rad} / \mathrm{s} $

Therefore, the angular frequency of the system in SI units is 30 . Hence, the correct option is (C).

Initially, the block is floating. The weight of the block is balanced by the buoyant force $\left(\mathrm{F}_{\mathrm{B}}\right)$.

Let h be the height of the submerged part of the block.

$ \mathrm{Mg}=\mathrm{F}_{\mathrm{B}}=\mathrm{V}_{\text {submerged }} \times \rho_{\text {liquid }} \times \mathrm{g} $

$\Rightarrow $ $\mathrm{Mg}=(\mathrm{A} \times \mathrm{h}) \rho \mathrm{g}$

When the block is depressed by a small distance x downwards, the volume of the displaced liquid increases by $A \cdot x$. This creates an extra buoyant force which acts as the restoring force.

The additional buoyant force is :

$ \mathrm{F}_{\text {restoring }}=-(\text { Additional Volume }) \times \rho \times \mathrm{g} $

$ \mathrm{F}_{\text {restoring }}=-(\mathrm{Ax}) \rho \mathrm{g} $

Since the force is proportional to displacement x and acts opposite to it ( $\mathrm{F} \propto-\mathrm{x}$ ), the motion is Simple Harmonic Motion (SHM).

Using Newton's Second Law ( $\mathrm{F}=\mathrm{Ma}$ ) :

$ \mathrm{Ma}=-(\rho \mathrm{Ag}) \mathrm{x} $

$\Rightarrow $ $ a=-(\rho A g M) x . $

Standard equation for SHM acceleration is $\mathrm{a}=-\omega^2 \mathrm{x}$

Comparing the two equations:

$ \omega^2=\frac{\rho \mathrm{Ag}}{\mathrm{M}} \Rightarrow \omega=\sqrt{\frac{\rho \mathrm{Ag}}{\mathrm{M}}} $

The time period T is given by $\mathrm{T}=\frac{2 \pi}{\omega}$.

$\Rightarrow $ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\rho \mathrm{Ag}}}$

Hence, the correct option is (C).

The spring constant k of a spring is inversely proportional to the length L of the spring.

$ \mathrm{k} \propto \frac{1}{\mathrm{~L}} $

This means that the product of the spring constant and the length is a constant for a given spring:

$ \mathrm{k} \cdot \mathrm{~L}=\text { Constant } $

The force constant of original spring is $\mathrm{k}_{\mathrm{o}}=15 \mathrm{~N} / \mathrm{m}$.

Let the original length be L .

The spring is cut into two pieces with a length ratio of $1: 3$.

$ x+3 x=L \Rightarrow x=\frac{L}{4} $

Hence, the length of smaller spring is $L_1=x=\frac{L}{4}$ and the length of larger spring is $L_2=3 x=\frac{3 L}{4}$.

Let $\mathrm{k}_1$ be the force constant of the smaller piece.

Using the inverse proportionality relationship:

$ \mathrm{k}_{\mathrm{o}} \mathrm{~L}_{\mathrm{o}}=\mathrm{k}_1 \mathrm{~L}_1 $

$\Rightarrow $ $ 15 \times \mathrm{L}=\mathrm{k}_1 \times\left(\frac{\mathrm{L}}{4}\right) \Rightarrow \mathrm{k}_1=60 \mathrm{~N} / \mathrm{m} $

Therefore, the force constant of smaller piece is $60 \mathrm{~N} / \mathrm{m}$.

Hence, the correct option is (C).

The time period (T) of a simple pendulum is the time taken to complete one full oscillation.

For a simple pendulum of the length of the string $(\mathrm{L})$ and the acceleration due to gravity $(\mathrm{g})$ the formula for the time period is,

$ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}} $

So, the time period is directly proportional to the square root of the length $(T \propto \sqrt{L})$.

$ \mathrm{T}^2 \propto \mathrm{~L} \Rightarrow \propto \mathrm{~T}^2 $

Case - 1

The initial length of the string is $\mathrm{L}_1=30 \mathrm{~cm}$

$\mathrm{N}_1=20$ is the number of oscillations completed in total time $\mathrm{t}_1=10 \mathrm{~s}$

So, the initial time period of oscillation is

$ \mathrm{T}_1=\frac{\text { Total Time }}{\text { Number of Oscillations }}=\frac{10}{20}=0.5 \mathrm{~s} $

Case - 2

$\mathrm{N}_2=40$ is the number of oscillations completed in the same time $\mathrm{t}_2=\mathrm{t}_1=10 \mathrm{~s}$

So, the final time period of oscillation is,

$ \mathrm{T}_2=\frac{\text { Total Time }}{\text { Number of Oscillations }}=\frac{10}{40}=0.25 \mathrm{~s} $

We need to find the required length $\left(\mathrm{L}_2\right)$ corresponding to the final length of the string, Using the proportionality $\mathrm{L} \propto \mathrm{T}^2$

$ \frac{\mathrm{L}_2}{\mathrm{~L}_1}=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2 $

$\Rightarrow $ $\frac{\mathrm{L}_2}{30}=\left(\frac{0.25}{0.50}\right)^2=\frac{1}{4}$

$\Rightarrow $ $ \frac{\mathrm{L}_2}{30}=\frac{1}{4} \Rightarrow \mathrm{~L}_2=\frac{30}{4} \mathrm{~cm}=7.5 \mathrm{~cm} $

Therefore, the required length of the string is 7.5 cm .

Hence, the correct option is (B).

The time period $(T)$ of a simple pendulum of length $L$ is given by the standard formula :

$ \mathrm{T}=2 \pi \sqrt{\mathrm{~L} / \mathrm{g}} $

Where $g$ is the acceleration due to gravity.

The graphs in the options plot $\frac{1}{\mathrm{~T}^2}$ versus L .

On rearranging the formula for the time period,

$ \mathrm{T}^2=\frac{4 \pi^2 \mathrm{~L}}{\mathrm{~g}} $

$\Rightarrow $ $\frac{1}{\mathrm{~T}^2}=\frac{\mathrm{g}}{4 \pi^2 \mathrm{~L}}$

$\Rightarrow $$ \frac{1}{\mathrm{~T}^2}=\left(\frac{\mathrm{g}}{4 \pi^2}\right) \frac{1}{\mathrm{~L}} $

Since $g$ and $4 \pi^2$ are constants, the term $\frac{g}{4 \pi^2}$ is just a constant value. Let $\frac{g}{4 \pi^2}=k$, where $k$ is any positive constant.

Therefore, the relationship is :

$ \mathrm{y}=\mathrm{kx} $

where $\mathrm{y}=\frac{1}{\mathrm{~T}^2}$ and $\mathrm{x}=\mathrm{L}$.

The mathematical equation $\mathrm{y}=\mathrm{kx}$ (or $\mathrm{xy}=$ constant) represents a rectangular hyperbola, where $\frac{1}{\mathrm{~T}^2} \rightarrow 0$ if $\mathrm{L} \rightarrow \infty$, i.e., x -axis is horizontal asymptote. Also, $\frac{1}{\mathrm{~T}^2} \rightarrow \infty$ if $\mathrm{L} \rightarrow 0$, i.e., y-axis is horizontal asymptote

So, the resulting graph is,

Therefore, the correct option is (A).

Let the displacement $(\mathrm{x})$ of a simple harmonic oscillator at any time t be represented by :

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

Where :

• A is the amplitude.

• $\omega$ is the angular frequency of the oscillator.

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

$ v=\frac{d x}{d t}=A \omega \cos (\omega t) $

The kinetic energy (K.E.) of the oscillator is given by :

$ \text { K. } \mathrm{E}=\frac{1}{2} \mathrm{~m} \mathrm{v}^2 $

$\Rightarrow $ K. $\mathrm{E} .=\frac{1}{2} \mathrm{~m}(\mathrm{~A} \omega \cos (\omega \mathrm{t}))^2$

$\Rightarrow $ K. $\mathrm{E} .=\frac{1}{2} \mathrm{~m} \mathrm{~A}^2 \omega^2 \cos ^2(\omega \mathrm{t})$

$\Rightarrow $ K. $E \cdot=\frac{1}{2} m A^2 \omega^2\left[\frac{1+\cos (2 \omega t)}{2}\right]$

$\Rightarrow $ K. $E \cdot=\frac{1}{4} m A^2 \omega^2+\frac{1}{4} m A^2 \omega^2 \cos (2 \omega t)$

$\Rightarrow $ $ \text { K. } \mathrm{E} .=\frac{1}{4} \mathrm{~m} \mathrm{~A}^2 \omega^2+\frac{1}{4} \mathrm{~m} \mathrm{~A}^2 \omega^2 \cos \left(\omega^{\prime} \mathrm{t}\right) $

Here, $\omega^{\prime}=2 \omega$ is the frequency of oscillation of kinetic energy of oscillator.

So, the kinetic energy oscillates with an angular frequency of $2 \omega$.

The kinetic energy oscillates with an angular frequency $\omega^{\prime}=176 \mathrm{rad} / \mathrm{s}$.

$ \Rightarrow \omega=\frac{176}{2}=88 \mathrm{rad} / \mathrm{s} $

Using the relationship between angular frequency ( $\omega$ ) and frequency (f) in Hz is:

$ \omega=2 \pi \mathrm{f} $

$\Rightarrow $ $ f=\frac{\omega}{2 \pi} $

Substitute the values $\left(\omega=88\right.$ and $\left.\pi=\frac{22}{7}\right)$ :

$ \mathrm{f}=\frac{88}{2 \times 22 / 7} $

$\Rightarrow $ $f=\frac{88 \times 7}{44}$

$\Rightarrow $ $\mathrm{f}=14 \mathrm{~Hz}$

Given:

Natural length of the spring: 2 meters

Initial compression of the spring ($ x_i $): 1 meter

Final compression of the spring when the block is at distance $ x $ ($ x_f $): $ (2 - x) $ meters

Solution:

To find the block's speed, we'll use energy conservation. Initially, the spring's potential energy is fully converted to kinetic energy and spring potential energy at the new position $ x $.

Conservation of Energy:

Initial kinetic energy ($ K_i $) is 0 since the block starts from rest.

Initial potential energy ($ U_i $) due to spring compression is $\frac{1}{2} K x_i^2$.

At a distance $ x $, the kinetic energy ($ K_f $) is $\frac{1}{2} mv^2$ and the potential energy ($ U_f $) is $\frac{1}{2} K x_f^2$.

$ K_i + U_i = K_f + U_f $

$ 0 + \frac{1}{2} K x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} K x_f^2 $

Rearrange to find $ v^2 $:

$ \frac{1}{2} m v^2 = \frac{1}{2} K (x_i^2 - x_f^2) $

Substitute known values:

$ \frac{1}{2} \times 2 \times v^2 = \frac{1}{2} \times 200 \times \left(1^2 - (2-x)^2 \right) $

Simplify and solve for $ v^2 $:

$ v^2 = 100 \left[ 1 - (2-x)^2 \right] $

Finally, take the square root to find $ v $:

$ v = 10 \left[ 1 - (2-x)^2 \right]^{1/2} $

Therefore, the speed of the block at distance $ x $ from the wall is given by the expression above.

$\begin{aligned} &\text { (A) As both blocks moving together so }\\ &\begin{aligned} & \text { Time period }=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} ; \text { where } \mathrm{m}=\mathrm{M}+\mathrm{m} \\ & \qquad \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{~K}}} \end{aligned} \end{aligned}$

(B) Let block is displaced by x in $(+\mathrm{ve})$ direction so force on block will be in(-ve) direction

$\begin{aligned} & F=-K x \\ & (M+m) a=-K x \end{aligned}$

$a=-\frac{K x}{(M+m)}$

(C) As upper block is moving due to friction thus

$\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mKx}}{(\mathrm{M}+\mathrm{m})}$

(D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together.

$\begin{aligned} & K A=(M+m) a \\ & a=\frac{K A}{(M+m)} \\ & f=m a=\frac{m K A}{(M+m)} \\ & f_{\max }=f_L=\mu m g \\ & f=\mu m g \\ & \frac{m K A}{(M+m)}=\mu m g \\ & A=\frac{\mu(M+m) g}{K} \end{aligned}$

(E) Maximum friction can be $\mu \mathrm{mg}$ as force is acting between blocks & normal force here is mg .

$\begin{aligned} &\begin{aligned} & x_1=\sqrt{7} \sin 5 t \\ & x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \end{aligned}\\ &\text { From phasor, } \end{aligned}$

$\begin{aligned} &\therefore \text { Amplitude of resultant } \mathrm{SHM}=7\\ &\begin{aligned} & \phi=\tan ^{-1} \frac{2 \sqrt{7} \times \sqrt{3} / 2}{\sqrt{7}+2 \sqrt{7} \times \frac{1}{2}}=\tan ^{-1} \frac{\sqrt{21}}{2 \sqrt{7}}=\tan ^{-1} \frac{\sqrt{3}}{2} \\ & \therefore X_R=7 \sin (5 \mathrm{t}+\phi) \\ & \quad a_R=-7 \times 25 \sin (5 \mathrm{t}+\phi) \\ & \therefore a_{\max }=175 \mathrm{~cm} / \mathrm{sec}=175 \times 10^{-2} \mathrm{~m} / \mathrm{sec} \end{aligned} \end{aligned}$

We know, for an oscillation,

${V_{\max }} = wA$ .... (1)

where, A = amplitude

w = frequency of oscillation

and $w = \sqrt {{k \over m}} $ .... (2)

from (1) and (2), ${V_{\max }} = \sqrt {{k \over m}} A$

$ \Rightarrow {V_{\max }} \propto \sqrt k $ (As m and A are same (or constant))

$ \Rightarrow {{{{\left( {{V_{\max }}} \right)}_A}} \over {{{\left( {{V_{\max }}} \right)}_B}}} = \sqrt {{{{k_1}} \over {{k_2}}}} $

Hence, option 2 is correct.

$\begin{aligned} &\text { As } h \text { increases, } g \text { decreases, } \mathrm{T} \text { increases }\\ &\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\ & \mathrm{~g}=\frac{\mathrm{g}_0 \mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \end{aligned} \end{aligned}$

$\begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \\ & \mathrm{x}_0=\mathrm{A} \sin \phi \quad\text{.... (1)}\\ & \mathrm{p}=\mathrm{mA} \omega \cos (\omega \mathrm{t}+\phi) \\ & \mathrm{p}_0=\mathrm{mA} \omega \cos \phi \quad\text{.... (2)} \end{aligned}$

$(2) /(1) \Rightarrow \tan \phi=\left(\frac{x_0}{p_0}\right) m \omega$

$\sin \phi=\frac{\mathrm{x}_0 \mathrm{~m} \omega}{\sqrt{\left(\mathrm{~m} \omega \mathrm{x}_0\right)^2+\mathrm{p}_0^2}}$

From (1), $A=\frac{x_0}{\sin \phi}=\frac{\sqrt{\left(m \omega x_0\right)^2+p_0^2}}{m \omega}$

This means we can explain assertion with the given reason.

$\begin{aligned} & x=x_0 \sin ^2 \frac{t}{2}=x_0\left(\frac{1-\cos t}{2}\right) \\ & x-\frac{x_0}{2}=\frac{-\cos t}{2} \\ & \text { where } x_0=1 \\ & x-\frac{1}{2}=\frac{-\cos t}{2} \end{aligned}$

Particle is oscillating between $\mathrm{x}=0$ to $\mathrm{x}=1$

$ x(t) = \cos(\pi t) $

The particle executes simple harmonic motion (SHM) with amplitude $ A = 1 \text{ cm} $ and period $ T = 2 \text{ s} $. In one complete cycle (2 s), the motion is as follows:

It starts at $ x = 1 \text{ cm} $.

It moves to $ x = -1 \text{ cm} $ (covering a distance of $ 2 \text{ cm} $).

It returns to $ x = 1 \text{ cm} $ (covering another $ 2 \text{ cm} $).

Thus, the total distance traveled in one complete cycle is:

$ D_{\text{cycle}} = 2 + 2 = 4 \text{ cm}. $

In 12.5 s, the number of cycles is:

$ \frac{12.5}{2} = 6.25 \text{ cycles}. $

For the 6 complete cycles (12 s), the distance traveled is:

$ D_{\text{complete}} = 6 \times 4 = 24 \text{ cm}. $

For the remaining 0.5 s, determine the displacement. At $ t = 12 \text{ s} $, the particle is at:

$ x(12) = \cos(12\pi) = 1 \text{ cm}. $

After an extra 0.5 s (i.e., at $ t = 12.5 \text{ s} $), the position becomes:

$ x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \text{ cm}. $

The distance covered in this interval is:

$ D_{\text{extra}} = |1 - 0| = 1 \text{ cm}. $

So, the total distance traveled is:

$ D = D_{\text{complete}} + D_{\text{extra}} = 24 + 1 = 25 \text{ cm}. $

The net displacement, which is the difference between the final and initial positions, is calculated as follows:

Initial position at $ t = 0 $:

$ x(0) = \cos 0 = 1 \text{ cm}, $

Final position at $ t = 12.5 \text{ s} $:

$ x(12.5) = 0 \text{ cm}. $

Thus, the displacement is:

$ d = |0 - 1| = 1 \text{ cm}. $

The ratio of the total distance traveled to the displacement is:

$ \frac{D}{d} = \frac{25}{1} = 25. $

Therefore, the correct answer is $25$.

We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force. Here’s a step‐by‐step explanation:

Define the given values:

Side length of the cube, $L = 10\text{ cm} = 0.1\text{ m}$.

Mass of the cube, $m = 10\text{ g} = 0.01\text{ kg}$.

Density of water, $\rho = 10^3\ \text{kg/m}^3$.

Acceleration due to gravity, $g = 10\ \text{m/s}^2$.

Cross-sectional area of the cube (face area),

$A = L^2 = (0.1)^2 = 0.01\text{ m}^2.$

When the cube is depressed by a small distance $x$, the additional volume of water displaced is

$\Delta V = A\,x.$

Thus, the additional buoyant force is:

$F_b = \rho\,g\,\Delta V = \rho\,g\,A\,x.$

This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement $x$, which is the hallmark of simple harmonic motion (SHM).

The effective spring constant, $k$, associated with this SHM is:

$k = \rho\,g\,A.$

The period of oscillation for SHM is given by:

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

which, upon substituting for $k$, becomes:

$T = 2\pi \sqrt{\frac{m}{\rho\,g\,A}}.$

Substitute the given values into the expression:

$T = 2\pi \sqrt{\frac{0.01}{1000 \times 10 \times 0.01}}.$

Calculate the denominator:

$1000 \times 10 \times 0.01 = 100,$

so,

$T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.$

Since,

$\sqrt{0.0001} = 0.01,$

we have:

$T = 2\pi \times 0.01 = 0.02\pi\ \text{s}.$

The problem states that the time period is given by:

$y \pi \times 10^{-2}\ \text{s}.$

Writing our result in the same form:

$0.02\pi\ \text{s} = 2\pi \times 10^{-2}\ \text{s}.$

We see that $y = 2.$

Therefore, the correct answer is:

Option A: 2.

Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times and 2 times that of the Earth, respectively, has the same time period as it does on Earth.

Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other planet.

Explanation:

The acceleration due to gravity $ g $ on a planet is given by the formula:

$ g = \frac{G M}{R^2} $

where $ G $ is the gravitational constant, $ M $ is the mass of the planet, and $ R $ is the radius of the planet.

For the new planet, the gravitational acceleration $ g' $ is:

$ g' = \frac{G(4M)}{(2R)^2} = \frac{4G M}{4R^2} = \frac{G M}{R^2} = g $

Thus, the gravitational acceleration on this new planet is the same as on Earth, $ g $.

The time period $ T $ of a simple pendulum is determined by:

$ T = 2\pi \sqrt{\frac{\ell}{g}} $

where $ \ell $ is the length of the pendulum, which indicates that the time period $ T $ is independent of the mass of the pendulum.

Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly explain Assertion (A) because the time period of the pendulum is also independent of the pendulum's mass, and the key factor is the unchanged gravitational acceleration.

The time period of a simple pendulum is given by the formula:

$T = 2\pi \sqrt{\frac{l}{g}}$

where $T$ is the time period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $h$ above the Earth's surface can be expressed as:

$g' = g \left(\frac{R}{R + h}\right)^2$

where $g$ is the acceleration due to gravity at the surface of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $g$ due to a change in height will affect the time period.

Given that the time period of the pendulum at a height $R$ above Earth's surface is $T_1$, and we're to find the time period $T_2$ at a height of $2R$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $T_1$ and $T_2$.

For the initial case at height $R$:

$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$

For the new case at height $2R$:

$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

The time period is proportional to the square root of the inverse of $g$, so:

$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Therefore:

$T_1 = \frac{2}{3}T_2$

Rearranging this equation:

$3T_1 = 2T_2$

This corresponds to Option A.

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

Here energy depends on $A$ and $k$ and not on mass.

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :

$ p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} $

After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $ v' $. The final momentum $ p_{\text{final}} $ is the combined mass of the bullet and bob times the common velocity :

$ p_{\text{final}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v' $

According to the principle of conservation of linear momentum,

$ p_{\text{initial}} = p_{\text{final}} $

$ m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v' $

Plugging in the values :

$ (10^{-2} \text{ kg}) \times (2 \times 10^2 \text{ m/s}) = (10^{-2} \text{ kg} + 1 \text{ kg}) \times v' $

Solving for $ v' $ :

$ v' = \frac{10^{-2} \times 2 \times 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 \text{ m/s} $

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $ h $ that the bob reaches. Using the principle of conservation of energy :

$ \text{Kinetic Energy (KE)}_{\text{initial}} = \text{Potential Energy (PE)}_{\text{final}} $

$ \frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2 = (m_{\text{bullet}} + m_{\text{bob}})gh $

Isolating $ h $, we get :

$ h = \frac{\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2}{(m_{\text{bullet}} + m_{\text{bob}})g} = \frac{v'^2}{2g} $

Substituting the values for $ v' $ and $ g $ :

$ h = \frac{(1.98)^2}{2 \times 10} = \frac{3.9204}{20} = 0.19602 \text{ m} $

Looking at the given options, the result most closely matches Option A :

$ \boxed{0.20 \text{ m}} $

Therefore, the height to which the bob rises before swinging back is approximately $ 0.20 \text{ m} $.

$\ell=10 \mathrm{~m}$,

Initial energy $=\mathrm{mg} \ell$

$\begin{aligned} & \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \\ & \mathrm{v}^2=180 \\ & \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{aligned}$

Let's denote the amplitude of the simple harmonic motion as A, and the displacement of the particle from the mean position as x. The given condition is that x = A/2.

For a particle in SHM, the potential energy (PE) is given by:

$PE = \frac{1}{2} kx^2$

And the total mechanical energy (E) of the particle remains constant and is given by:

$E = \frac{1}{2} kA^2$

Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:

$E = PE + KE$

Now, we need to find the ratio of potential energy to kinetic energy when x = A/2.

Calculate the potential energy at x = A/2:

$PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2$

Substitute the expression for total mechanical energy:

$KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2$

Now, find the ratio of potential energy to kinetic energy:

$\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$

Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$.

If we substitute these initial conditions into the equation for the displacement of the particle:

$x = A \sin(wt + \delta)$

We have:

$\frac{A}{2} = A \sin(\delta)$

Dividing both sides by $A$ gives us:

$\frac{1}{2} = \sin(\delta)$

The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees).

Therefore, the correct answer is $\delta = \frac{\pi}{6}$.

$\omega = \sqrt {{K \over m}} \Rightarrow \omega \propto {1 \over {\sqrt m }}$

${{{\omega _2}} \over {{\omega _1}}} = \sqrt {{{{m_1}} \over {{m_2}}}} = \sqrt {{1 \over 2}} $

$\left| {{g_{eff}}} \right| = \left| {\overline g - \overline a } \right|$

$ \Rightarrow {g_{eff}} = g\cos \theta $

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

$ = 2\pi = \sqrt {{L \over {g\cos \theta }}} $

$T \propto \sqrt {1/g} $

$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{g_2}} \over {{g_1}}}} = {R \over {R + h}}$

${4 \over 6} = {R \over {R + h}}$

$ \Rightarrow h = R/2$

$ = 3200$ km

Let $x = A\sin \omega t$

$ \Rightarrow v = A\omega \cos \omega t$

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

$ \Rightarrow {{{v^2}} \over {{\omega ^2}}} + {x^2} = {A^2}$

$\Rightarrow$ Ellipse

Both the springs are in parallel combination in both the diagrams so

${T_1} = 2\pi \sqrt {{{3m} \over {2k}}} $

and ${T_2} = 2\pi \sqrt {{m \over {3k}}} $

So, ${{{T_1}} \over {{T_2}}} = {3 \over {\sqrt 2 }}$

$\omega = \pi = \sqrt {{g \over l}} $

So, $l = {g \over {{\pi ^2}}}$

$ \simeq 99.4$ cm (Nearest value)

$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)$

$ = 4\cos (\omega t)$

$y = 4\sin (\omega t)$

$ \Rightarrow {x^2} + {y^2} = {4^2}$

$\Rightarrow$ The particle is moving in a circular motion with radius of 4 m.

$x = \sin \left( {\pi t + {\pi \over 3}} \right)m$

$ \Rightarrow {{dx} \over {dt}} = \pi \cos \left( {\pi t + {\pi \over 3}} \right)$

$ = \pi \cos \left( {\pi + {\pi \over 3}} \right)$ at $t = 1\,s$

$ = - {\pi \over 2}$ m/s

or $\left| {{{dx} \over {dt}}} \right| = 157$ cm/s

Time taken by the harmonic oscillator to move from mean position to half of amplitude is ${T \over {12}}$

So, ${T \over {12}}$ = 3

T = 36 sec.

${\omega _1}{A_1} = {\omega _2}{A_2}$

$ \Rightarrow {{{A_1}} \over {{A_2}}} = {{{\omega _2}} \over {{\omega _1}}}$

$ = \sqrt {{{{k_2}} \over {{m_2}}}} \times \sqrt {{{{m_1}} \over {{k_1}}}} = \sqrt {{{9k} \over {100}} \times {{50} \over {2k}}} = {3 \over 2}$