SHM of Simple Pendulum Questions in English

(C) Let $T_1$ and $T_2$ be the time periods of the two pendulums,and $\ell_1$ and $\ell_2$ be their lengths.
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.
Given that in the same time $t$,the first pendulum makes $n_1 = 11$ oscillations and the second makes $n_2 = 9$ oscillations.
Thus,$t = n_1 T_1 = n_2 T_2$.
$11 \times 2\pi \sqrt{\frac{\ell_1}{g}} = 9 \times 2\pi \sqrt{\frac{\ell_2}{g}}$.
Squaring both sides,we get $121 \times \frac{\ell_1}{g} = 81 \times \frac{\ell_2}{g}$.
Therefore,$\frac{\ell_1}{\ell_2} = \frac{81}{121}$.

(C) The radius of the circular path in the horizontal plane is $r = l \sin \theta$.
The forces acting on the bob are:
$(i)$ $T$,the tension in the string.
$(ii)$ $Mg$,the weight of the bob acting downwards.
Resolving the tension $T$ into vertical and horizontal components:
Vertical component: $T \cos \theta = Mg$ --- $(1)$
Horizontal component (centripetal force): $T \sin \theta = Mr \omega^2 = M(l \sin \theta) \omega^2$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{T \sin \theta}{T \cos \theta} = \frac{M(l \sin \theta) \omega^2}{Mg}$
$\tan \theta = \frac{l \sin \theta \cdot \omega^2}{g}$
$\frac{\sin \theta}{\cos \theta} = \frac{l \sin \theta \cdot \omega^2}{g}$
$\frac{1}{\cos \theta} = \frac{l \omega^2}{g}$
$\omega^2 = \frac{g}{l \cos \theta}$
$\omega = \sqrt{\frac{g}{l \cos \theta}}$
The time period $t$ is given by $t = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$:
$t = 2 \pi \sqrt{\frac{l \cos \theta}{g}}$

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
This implies $T \propto \sqrt{L}$.
Let the initial length be $L_1 = L$ and the initial time period be $T_1 = T$.
The new length is $L_2 = L + 0.44L = 1.44L$.
The new time period $T_2$ is given by $\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}$.
Substituting the values,$\frac{T_2}{T} = \sqrt{\frac{1.44L}{L}} = \sqrt{1.44} = 1.2$.
Thus,$T_2 = 1.2T$.
The change in time period is $\Delta T = T_2 - T_1 = 1.2T - T = 0.2T$.
The percentage increase is $\frac{\Delta T}{T_1} \times 100 = \frac{0.2T}{T} \times 100 = 20\%$.

(B) The frequency of a simple pendulum is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.
Given the frequency relation $f_{A}=2 f_{B}$.
Substituting the formula for frequency:
$\frac{1}{2 \pi} \sqrt{\frac{g}{l_{A}}} = 2 \times \frac{1}{2 \pi} \sqrt{\frac{g}{l_{B}}}$
Squaring both sides:
$\frac{g}{l_{A}} = 4 \times \frac{g}{l_{B}}$
$\frac{1}{l_{A}} = \frac{4}{l_{B}}$
Therefore,$l_{A} = \frac{l_{B}}{4}$.
Since the frequency of a simple pendulum depends only on the length $l$ and acceleration due to gravity $g$,it does not depend on the mass $M$ of the bob.

(A) Let $O$ be the point of suspension and $B$ be the equilibrium position. The bob is displaced to point $C$ at an angle $\theta$ with the vertical.
In $\triangle OAC$,the vertical distance $OA = l \cos \theta$.
The vertical height $h$ through which the bob falls from $C$ to $B$ is $h = OB - OA = l - l \cos \theta = l(1 - \cos \theta)$.
By the law of conservation of energy,the potential energy lost equals the kinetic energy gained at the equilibrium position $B$:
$mgh = \frac{1}{2}mv^2$
$v^2 = 2gh$
Substituting $h = l(1 - \cos \theta)$:
$v^2 = 2gl(1 - \cos \theta)$
$v = \sqrt{2gl(1 - \cos \theta)}$

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.
When the train accelerates with $a = 10 \ m/s^2$,the effective gravity is $g_{eff} = \sqrt{g^2 + a^2}$.
When the train retards with $a = 10 \ m/s^2$,the effective gravity is $g'_{eff} = \sqrt{g^2 + (-a)^2} = \sqrt{g^2 + a^2}$.
Since the magnitude of the effective acceleration $g_{eff}$ remains the same in both cases,the time period $T$ remains unchanged.
Therefore,the time period remains $2 \ s$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.
When an elevator moves with constant velocity (constant speed in a straight line),its acceleration is zero.
In this case,the effective acceleration $g_{eff} = g$,which is the same as when the elevator is at rest.
Therefore,the time period $T$ remains unchanged,and the clock shows the correct time.
If the elevator accelerates,$g_{eff}$ changes $(g_{eff} = g \pm a)$,causing the clock to gain or lose time.

(C) The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the bob is immersed in a liquid,it experiences an upward buoyant force $F_{up} = V \sigma g$,where $V$ is the volume of the bob and $\sigma$ is the density of the liquid.
The effective weight of the bob becomes $mg' = mg - F_{up} = V \rho_0 g - V \sigma g$,where $\rho_0$ is the density of the bob material.
Thus,the effective acceleration due to gravity $g'$ is $g' = g \left(1 - \frac{\sigma}{\rho_0}\right)$.
Given the ratio $\rho = \frac{\sigma}{\rho_0}$,we have $g' = g(1 - \rho)$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g(1 - \rho)}}$.
Comparing this with the original time period $T$,we get $T' = \frac{T}{\sqrt{1 - \rho}}$.

(A) For a simple harmonic oscillator,the potential energy $V$ at displacement $x$ is given by $V = \frac{1}{2} k x^2$,where $k = m \omega^2$.
The kinetic energy $T$ at displacement $x$ is given by $T = \frac{1}{2} k (A^2 - x^2)$.
Taking the ratio $\frac{V}{T}$:
$\frac{V}{T} = \frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)}$
$\frac{V}{T} = \frac{x^2}{A^2 - x^2}$

(A) The displacement of the bob in simple harmonic motion is given by $x = A \sin(\frac{2\pi}{T}t)$.
In one complete oscillation (time $T$),the bob travels from the mean position to the extreme position $A$,back to the mean position,to the other extreme position $-A$,and returns to the mean position.
The total distance covered in one time period $T$ is $A + A + A + A = 4A$.
The average speed is defined as the total distance divided by the total time.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{4A}{T}$.

(D) The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.
When an elevator accelerates downwards with an acceleration $a$,the effective acceleration becomes $g_{eff} = (g - a)$.
Since $T \propto \frac{1}{\sqrt{g_{eff}}}$,as $g_{eff}$ decreases,the time period $T$ increases.
Among the given options,when the elevator is accelerating downwards,$g_{eff}$ is minimized (assuming $a < g$),resulting in the greatest time period.

(A) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$.
Here,$L$ is the effective length of the pendulum and $g$ is the acceleration due to gravity.
As observed from the formula,the time period $T$ is independent of the mass of the bob.
Since the length of the pendulum and the acceleration due to gravity remain unchanged,the time period will remain the same regardless of the change in mass.
Therefore,the new time period is $2\, s$.

(B) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l_{eff}}{g}}$.
When the child is sitting,the center of mass of the system is at a certain distance from the point of suspension.
When the child stands up,the center of mass shifts upwards,which effectively decreases the length of the pendulum $(l_{eff})$.
Since $T \propto \sqrt{l_{eff}}$,a decrease in $l_{eff}$ results in a decrease in the time period $T$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L_{eff}}{g}}$,where $L_{eff}$ is the distance between the point of suspension and the center of mass of the bob.
Initially,the center of mass of the water-filled sphere is at its geometric center.
As water starts leaking out from the bottom,the center of mass of the remaining water shifts downwards. This increases the effective length $L_{eff}$ of the pendulum.
Since $T \propto \sqrt{L_{eff}}$,the time period $T$ increases as the center of mass moves down.
When the water is completely drained,the center of mass returns to the geometric center of the sphere,so the effective length returns to its original value,and the time period decreases back to its original value.
Therefore,the time period first increases and then decreases to the original value.

(A) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$,where $g_{eff}$ is the effective acceleration due to gravity.
Inside an orbiting satellite,the body is in a state of weightlessness,which means the effective acceleration due to gravity $g_{eff} = 0$.
Substituting $g_{eff} = 0$ into the formula,we get $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
Thus,the Assertion is correct.
The Reason states that the time period is inversely proportional to $\sqrt{g}$,which is mathematically correct based on the formula $T \propto \frac{1}{\sqrt{g}}$.
Since the infinite time period is a direct consequence of $g_{eff} = 0$ in the formula,the Reason correctly explains the Assertion.

(D) The time of fall $t$ for an object falling from height $h$ is given by $t = \sqrt{\frac{2h}{g}}$. Thus,$t \propto \frac{1}{\sqrt{g}}$.
The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$. Thus,$T \propto \frac{1}{\sqrt{g}}$.
Taking the ratio,we get $\frac{t}{T} = \frac{k_1 / \sqrt{g}}{k_2 / \sqrt{g}} = \text{constant}$.
Since the ratio $\frac{t}{T}$ is independent of the acceleration due to gravity $g$,the relationship remains the same on any planet.
Given that on Earth $t = 2T$,it follows that on the other planet $t' = 2T'$.

(B) The time period of a simple pendulum is given by the formula:
$T = 2 \pi \sqrt{\frac{L}{g}}$
Rearranging the formula to solve for the length $L$:
$L = \frac{g T^{2}}{4 \pi^{2}}$
$A$ seconds pendulum is defined as a pendulum that has a time period of $T = 2 \, s$ (one second for each tick).
Given $g = 9.8 \, m/s^{2}$ and $T = 2 \, s$:
$L = \frac{9.8 \times (2)^{2}}{4 \times (3.14)^{2}}$
Since $\pi^{2} \approx 9.8$,we have:
$L = \frac{9.8 \times 4}{4 \times 9.8} = 1 \, m$
Thus,the length of the seconds pendulum is $1 \, m$.

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
For the earth: $T_e = 2 \pi \sqrt{\frac{l}{g_e}} = 3.5 \; s$,where $g_e = 9.8 \; m s^{-2}$.
For the moon: $T_m = 2 \pi \sqrt{\frac{l}{g_m}}$,where $g_m = 1.7 \; m s^{-2}$.
Taking the ratio: $\frac{T_m}{T_e} = \sqrt{\frac{g_e}{g_m}}$.
$T_m = T_e \times \sqrt{\frac{g_e}{g_m}} = 3.5 \times \sqrt{\frac{9.8}{1.7}}$.
$T_m = 3.5 \times \sqrt{5.7647} \approx 3.5 \times 2.4 = 8.4 \; s$.
Thus,the time period on the moon is $8.4 \; s$.

(N/A) For a simple pendulum,the restoring force is $F = -mg \sin \theta$. For small angles,$\sin \theta \approx \theta$,so $F \approx -mg \theta = -mg (x/l) = -(mg/l)x$. Comparing this with $F = -kx$,we get $k = mg/l$. Substituting this into $T = 2 \pi \sqrt{m/k}$,we get $T = 2 \pi \sqrt{m / (mg/l)} = 2 \pi \sqrt{l/g}$. Thus,$T$ is independent of mass $m$.
$(b)$ For larger angles,$\sin \theta < \theta$. The restoring force $F = -mg \sin \theta$ is smaller than the linear approximation $F = -mg \theta$. $A$ smaller restoring force leads to a slower motion and a longer time period $T$.
$(c)$ Yes. $A$ wristwatch works on the principle of spring-driven mechanical oscillations or quartz crystal vibrations,which are independent of the acceleration due to gravity $g$. Thus,it gives the correct time during free fall.
$(d)$ In a freely falling cabin,the effective acceleration due to gravity $g_{eff} = g - a = g - g = 0$. The time period $T = 2 \pi \sqrt{l/g_{eff}} = \infty$. Therefore,the frequency $f = 1/T = 0$.

(N/A) The bob of the simple pendulum experiences two accelerations: the acceleration due to gravity $(g)$ acting vertically downwards and the centripetal acceleration $(a_c = v^2/R)$ acting horizontally towards the center of the circular track.
The effective acceleration $(a_{\text{eff}})$ is the vector sum of these two perpendicular accelerations:
$a_{\text{eff}} = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + (v^2/R)^2}$
The time period $(T)$ of a simple pendulum is given by the formula:
$T = 2\pi \sqrt{\frac{l}{a_{\text{eff}}}}$
Substituting the value of $a_{\text{eff}}$:
$T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + (v^2/R)^2}}}$

(N/A) Base area of the cork $= A$
Height of the cork $= h$
Density of the liquid $= \rho_{l}$
Density of the cork $= \rho$
In equilibrium, the weight of the cork equals the weight of the liquid displaced by the floating cork.
Let the cork be depressed slightly by a distance $x$. As a result, an additional volume of liquid is displaced, creating an extra upward buoyant force (up-thrust) which acts as the restoring force.
Restoring force $F = -(\text{Weight of extra liquid displaced})$
$F = -(A \cdot x \cdot \rho_{l} \cdot g)$
According to the simple harmonic motion force law, $F = -kx$, where $k$ is the force constant.
Comparing the two expressions, $k = A \rho_{l} g$.
The time period of oscillation is given by $T = 2 \pi \sqrt{\frac{m}{k}}$, where $m$ is the mass of the cork.
Mass of the cork $m = \text{Volume} \times \text{Density} = (A \cdot h) \cdot \rho$.
Substituting $m$ and $k$ into the time period formula:
$T = 2 \pi \sqrt{\frac{A h \rho}{A \rho_{l} g}} = 2 \pi \sqrt{\frac{h \rho}{\rho_{l} g}}$.

(N/A) Simple pendulum: $A$ system consisting of a small,heavy body (bob) suspended by a light,inextensible,and flexible string from a fixed (rigid) support is called a simple pendulum.
The entire mass of the simple pendulum is considered to be concentrated at the center of gravity of the suspended bob.
The distance from the point of suspension to the center of mass of the bob is called the length of the pendulum $(L)$.
Derivation of the expression for the time period $(T)$:
Consider a simple pendulum with a small bob of mass $m$ tied to an inextensible,massless string of length $L$.
The other end of the string is fixed to a support. Let $\theta$ be the angle made by the string with the vertical.
Two forces act on the bob:
$(1)$ Tension $T$ along the string.
$(2)$ Gravitational force $mg$ acting vertically downwards.
The force $mg$ can be resolved into two components:
$(1)$ Radial component: $mg \cos \theta$ (along the string).
$(2)$ Tangential component: $mg \sin \theta$ (perpendicular to the string).
The restoring force is $F = -mg \sin \theta$. For small oscillations,$\sin \theta \approx \theta$ (in radians).
So,$F = -mg \theta = -mg (x/L)$,where $x$ is the displacement.
Comparing with $F = -kx$,we get $k = mg/L$.
The time period is $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{m / (mg/L)} = 2\pi \sqrt{L/g}$.

(N/A) simple pendulum is defined as an idealized system consisting of a heavy point mass (called a bob) suspended from a rigid support by a massless,inextensible,and perfectly flexible string.
When the bob is displaced from its equilibrium position and released,it undergoes periodic motion.
The length of the pendulum,denoted by $L$,is defined as the distance from the point of suspension to the center of gravity of the bob.

(N/A) simple pendulum consists of a bob of mass $m$ suspended by a light string of length $l$. When the pendulum is displaced by a small angle $\theta$ from the mean position,the restoring force is provided by the tangential component of gravity.
$1$. The restoring force is given by $F = -mg \sin \theta$.
$2$. For small angles,$\sin \theta \approx \theta$ (in radians),where $\theta = \frac{x}{l}$ ($x$ is the linear displacement).
$3$. Substituting this,we get $F = -mg \left( \frac{x}{l} \right) = -\left( \frac{mg}{l} \right) x$.
$4$. Since $m, g,$ and $l$ are constants,the restoring force $F$ is directly proportional to the negative displacement $x$,i.e.,$F \propto -x$.
$5$. This is the defining condition for Simple Harmonic Motion $(SHM)$. Thus,the motion is simple harmonic with an angular frequency $\omega = \sqrt{\frac{g}{l}}$.

(N/A) The laws of a simple pendulum are as follows:
$1$. Law of Isochronism: For small amplitudes,the period of oscillation $(T)$ of a simple pendulum is independent of its amplitude.
$2$. Law of Length: The period of oscillation $(T)$ is directly proportional to the square root of its effective length $(l)$,i.e.,$T \propto \sqrt{l}$.
$3$. Law of Acceleration due to Gravity: The period of oscillation $(T)$ is inversely proportional to the square root of the acceleration due to gravity $(g)$,i.e.,$T \propto 1/\sqrt{g}$.
$4$. Law of Mass: The period of oscillation $(T)$ is independent of the mass,shape,and material of the bob,provided the length remains constant.

(B) In a simple pendulum,the total mechanical energy is conserved.
At the mean position,the potential energy is minimum (taken as $0$),and the kinetic energy is maximum.
Since kinetic energy $K = \frac{1}{2}mv^2$,the speed $v$ is maximum when the kinetic energy is maximum.
Therefore,the speed of the pendulum is maximum at the mean position.

(B) No,the oscillation of a simple pendulum at the centre of the earth is not possible.
The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{l}{g}}$.
At the centre of the earth,the acceleration due to gravity $g$ is $0$.
Substituting $g = 0$ into the formula,we get $T = 2 \pi \sqrt{\frac{l}{0}}$,which results in $T = \infty$ (infinite).
Since the time period is infinite,the pendulum does not complete any oscillation.

(N/A) The time period of a simple pendulum (which models a swing) is given by the formula $T = 2 \pi \sqrt{\frac{l}{g}}$.
In this formula,$l$ represents the effective length of the swing (the distance from the point of suspension to the center of mass of the system) and $g$ is the acceleration due to gravity.
Since the mass of the person$(s)$ does not appear in the formula for the time period,the time period is independent of the mass of the person sitting on the swing.
Therefore,adding another person does not change the time period,provided the effective length $l$ remains constant.

(C) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.
From this formula, it is evident that the time period $T$ depends only on the length of the pendulum and the acceleration due to gravity at that location.
The time period is independent of the mass of the bob.
Therefore, if the mass of the bob is increased by $9$ times, the period of the pendulum will remain unchanged.

(A) The time period of a seconds pendulum is $T_{1} = 2 \ s$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Since $2\pi$ and $g$ are constants,we have the relation $T \propto \sqrt{l}$.
Given that the new length $l_{2} = \frac{l_{1}}{3}$,we can write the ratio:
$\frac{T_{2}}{T_{1}} = \sqrt{\frac{l_{2}}{l_{1}}} = \sqrt{\frac{l_{1}/3}{l_{1}}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.
Therefore,the new time period is $T_{2} = \frac{T_{1}}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ s$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
If the clock is running fast,it means the time period $T$ is smaller than required,causing it to complete more oscillations in a given time.
To adjust the clock to the true time,we need to increase the time period $T$.
Since $T \propto \sqrt{L}$,increasing the length $L$ of the pendulum will increase the time period $T$,thereby slowing down the clock to the correct time.

(B) The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{l}{g}}$,where $l$ is the effective length of the pendulum (distance from the point of suspension to the centre of gravity of the system).
When the girl is sitting,the centre of gravity is at a certain distance from the point of suspension.
When she stands up,the centre of gravity of the system shifts upwards,closer to the point of suspension.
This results in a decrease in the effective length $l$ of the pendulum.
Since $T \propto \sqrt{l}$,a decrease in $l$ leads to a decrease in the time period $T$.

(C) When two pendulums of equal length cross each other at the mean position,they are moving in opposite directions.
If one pendulum is at the mean position moving towards the right,its displacement is $x = A \sin(\omega t)$,which implies it is at phase $\phi_1 = 0$ or $\pi$.
If the other pendulum is crossing the same mean position moving towards the left,its motion is represented by $x = A \sin(\omega t + \pi)$.
The phase difference between the two motions is $\Delta \phi = \pi - 0 = \pi \text{ rad}$ or $180^{\circ}$.

(N/A) No,a simple pendulum clock cannot be used in an artificial satellite. The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$. In an artificial satellite,objects experience a state of weightlessness,which means the effective acceleration due to gravity $(g_{eff})$ is $0$. As $g_{eff} \to 0$,the time period $T \to \infty$. This implies that the pendulum will not oscillate,and the clock will stop working.

(B) The tension $T$ in the string of a simple pendulum at an angle $\theta$ with the vertical is given by $T = mg \cos \theta + \frac{mv^2}{l}$.
At the mean position,$\theta = 0^{\circ}$,so $\cos \theta = 1$ (maximum value).
Also,the velocity $v$ is maximum at the mean position.
Since both terms $mg \cos \theta$ and $\frac{mv^2}{l}$ are at their maximum values at the mean position,the total tension $T$ is maximum at the mean position.

(A) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
Since the acceleration due to gravity on the Moon $(g_m)$ is approximately $1/6$th of the acceleration due to gravity on Earth $(g_e)$,we have $g_m < g_e$.
According to the relation $T \propto \frac{1}{\sqrt{g}}$,as the value of $g$ decreases,the time period $T$ increases.
Therefore,when a simple pendulum is taken from the Earth's surface to the Moon's surface,its time period increases.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
$(a)$ For $T^2 \to l$: Squaring both sides, $T^2 = \frac{4\pi^2}{g} l$. Since $T^2 \propto l$, the graph is a straight line passing through the origin. Thus, $(a-i)$.
$(b)$ For $T^2 \to g$: From $T^2 = \frac{4\pi^2 l}{g}$, we have $T^2 \propto \frac{1}{g}$. This represents a rectangular hyperbola. Thus, $(b-iii)$.
$(c)$ For $T \to l$: From $T = 2\pi \sqrt{\frac{l}{g}}$, we have $T \propto \sqrt{l}$. This represents a parabolic curve. Thus, $(c-ii)$.
Therefore, the correct matching is $(a-i, b-iii, c-ii)$.

(B) The restoring force on a simple pendulum is given by $F = -mg \sin \theta$.
For small angular displacements,$\sin \theta \approx \theta$ (in radians).
Thus,the restoring force becomes $F \approx -mg \theta$.
Since $\theta = x/l$,we have $F \approx -(mg/l)x$.
This shows that the restoring force is directly proportional to the displacement $x$ and directed towards the equilibrium position,which is the condition for simple harmonic motion.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Since $T$ is constant $(2 \ s)$ for a second's pendulum on both the earth and the moon,we have $T \propto \sqrt{\frac{l}{g}}$.
Therefore,$\frac{T_m}{T_e} = \sqrt{\frac{l_m}{g_m} \cdot \frac{g_e}{l_e}}$.
Given $T_e = T_m = 2 \ s$,we get $1 = \sqrt{\frac{l_m}{l_e} \cdot \frac{g_e}{g_m}}$.
Squaring both sides,$1 = \frac{l_m}{l_e} \cdot \frac{g_e}{g_m}$.
We know that $g_m = \frac{g_e}{6}$,so $\frac{g_e}{g_m} = 6$.
Given $l_e = 1 \ m$,substituting these values gives $1 = \frac{l_m}{1} \cdot 6$.
Thus,$l_m = \frac{1}{6} \ m$.

(N/A) The angular frequency is $\omega = \frac{2\pi}{T} = 2\pi \text{ rad/s}$.
At $t=0$,the bob is at $\theta = \theta_0$. The angular position is $\theta(t) = \theta_0 \cos(\omega t)$.
At the moment the string snaps,$\theta = \frac{\theta_0}{2}$,so $\frac{\theta_0}{2} = \theta_0 \cos(2\pi t_1) \implies \cos(2\pi t_1) = \frac{1}{2} \implies 2\pi t_1 = \frac{\pi}{3} \implies t_1 = \frac{1}{6} \text{ s}$.
The velocity components at this instant are $v_x = l\omega \sin(\frac{\theta_0}{2})$ (horizontal) and $v_y = l\omega \cos(\frac{\theta_0}{2})$ (downward).
Since $\theta_0$ is small,$v_x \approx l\omega(\frac{\theta_0}{2})$ and $v_y \approx l\omega$.
The height of the bob from the ground at this instant is $h = H + l(1 - \cos(\frac{\theta_0}{2})) \approx H + l(1 - 1) = H$.
The vertical motion is $h = v_y t + \frac{1}{2}gt^2$,so $H = (l\omega)t + \frac{1}{2}gt^2$.
Solving for $t$ gives the time to hit the ground. The horizontal distance from $A$ is $x = l \sin(\frac{\theta_0}{2}) + v_x t = l(\frac{\theta_0}{2}) + (l\omega \frac{\theta_0}{2})t$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
In a freely falling lift,the effective acceleration due to gravity $g_{eff}$ is $g - a$. Since the lift is in free fall,$a = g$,therefore $g_{eff} = g - g = 0$.
Substituting this into the time period formula,we get $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
The frequency of oscillation $f$ is the reciprocal of the time period,$f = \frac{1}{T}$.
Therefore,$f = \frac{1}{\infty} = 0 \text{ Hz}$.

(D) When the lift is stationary,the time period of the simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
When the lift moves upwards with an acceleration $a = g/2$,a pseudo force acts downwards on the pendulum bob.
The effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{2} = \frac{3g}{2}$.
The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{L}{g_{eff}}}$.
Substituting $g_{eff} = \frac{3g}{2}$,we get $T' = 2 \pi \sqrt{\frac{L}{3g/2}} = 2 \pi \sqrt{\frac{2L}{3g}}$.
Comparing this with the original time period $T$,we have $T' = \sqrt{\frac{2}{3}} \times (2 \pi \sqrt{\frac{L}{g}}) = \sqrt{\frac{2}{3}} T$.

(C) The formula for the time period $T$ of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Given: Length $l = 2 \ m$ and time period $T = 2 \ s$.
Substituting the values into the formula:
$2 = 2 \pi \sqrt{\frac{2}{g}}$
Dividing both sides by $2$:
$1 = \pi \sqrt{\frac{2}{g}}$
Squaring both sides:
$1 = \pi^{2} \left(\frac{2}{g}\right)$
Rearranging for $g$:
$g = 2 \pi^{2} \ m/s^{2}$.

(B) second's pendulum is defined as a pendulum whose time period of oscillation is $2 \, s$.
Therefore,Statement $I$ is false.
The time taken to move from one extreme position to the other extreme position is equal to half of the time period $(T/2)$.
Since $T = 2 \, s$,the time taken is $2 \, s / 2 = 1 \, s$.
Therefore,Statement $II$ is true.

(A) Starting from the mean position,the particle moves to the extreme position ($1/4$ oscillation),back to the mean position ($1/2$ oscillation),and then to the other side.
$5/8$ of an oscillation is equivalent to $1/2 + 1/8$ of an oscillation.
In terms of phase,$1/2$ oscillation corresponds to a phase change of $\pi$ radians.
The remaining $1/8$ oscillation corresponds to a phase change of $\frac{1}{8} \times 2\pi = \frac{\pi}{4}$ radians.
However,the question asks for the time taken to reach the position corresponding to $5/8$ of the cycle. Using the reference circle method,the displacement $y = A \sin(\omega t)$.
For $5/8$ of the cycle,the phase angle $\phi = \frac{5}{8} \times 2\pi = \frac{5\pi}{4}$.
Since the particle starts from the mean position,we consider the phase relative to the mean position.
The time $t$ is given by $\omega t = \phi$,where $\omega = \frac{2\pi}{T}$.
Thus,$t = \frac{\phi}{\omega} = \frac{5\pi/4}{2\pi/T} = \frac{5}{8} T$.
Wait,checking the provided solution logic: The particle reaches the extreme at $T/4$,returns to mean at $T/2$. To complete $5/8$,it travels an additional $1/8$ of the cycle from the mean position in the negative direction. The displacement is $y = A \sin(\omega t)$. At $t = 5T/8$,$y = A \sin(2\pi/T \times 5T/8) = A \sin(5\pi/4) = -A/\sqrt{2}$.
If the question implies the time to reach the displacement corresponding to $5/8$ of the cycle,then $\alpha = 5$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.
Taking the logarithmic derivative,we get $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta \ell}{\ell}$.
Given that the length increases by $0.1\, \%$,so $\frac{\Delta \ell}{\ell} = \frac{0.1}{100} = 10^{-3}$.
The error in time per day $(\Delta T)$ is calculated as $\Delta T = \frac{1}{2} \times \left( \frac{\Delta \ell}{\ell} \right) \times T_{total}$,where $T_{total} = 24 \times 3600 \, s$.
Substituting the values: $\Delta T = \frac{1}{2} \times 10^{-3} \times 86400 = 43.2 \, s$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
When the bob is immersed in a liquid,the effective acceleration due to gravity $g_{\text{eff}}$ changes due to the buoyant force.
$m g_{\text{eff}} = m g - F_B$,where $F_B$ is the buoyant force.
$F_B = V \sigma g$,where $V$ is the volume of the bob and $\sigma$ is the density of the liquid.
Given $\sigma = \frac{\rho}{4}$,where $\rho$ is the density of the bob,we have $F_B = V \frac{\rho}{4} g = \frac{m g}{4}$.
Thus,$m g_{\text{eff}} = m g - \frac{m g}{4} = \frac{3 m g}{4}$,which implies $g_{\text{eff}} = \frac{3 g}{4}$.
The new length of the thread is $l_1 = l + \frac{l}{3} = \frac{4l}{3}$.
The new time period $T_1$ is given by $T_1 = 2 \pi \sqrt{\frac{l_1}{g_{\text{eff}}}} = 2 \pi \sqrt{\frac{4l/3}{3g/4}} = 2 \pi \sqrt{\frac{16l}{9g}} = \frac{4}{3} \left( 2 \pi \sqrt{\frac{l}{g}} \right)$.
Therefore,$T_1 = \frac{4}{3} T$.

(D) The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{\ell}{g}}$.
Initially,the time period is $T_{0} = 2 \pi \sqrt{\frac{\ell}{g}}$.
When the length is reduced to $\ell' = \frac{\ell}{16}$,the new time period $T'$ becomes:
$T' = 2 \pi \sqrt{\frac{\ell'}{g}} = 2 \pi \sqrt{\frac{\ell / 16}{g}}$.
$T' = \frac{1}{\sqrt{16}} \times (2 \pi \sqrt{\frac{\ell}{g}})$.
$T' = \frac{1}{4} T_{0}$.

(D) Given the equation for the frequency of a simple pendulum: $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$.
Squaring both sides,we get: $f^2 = \frac{1}{4 \pi^2} \cdot \frac{g}{\ell}$.
This can be rewritten as: $f^2 = \left( \frac{g}{4 \pi^2} \right) \cdot \left( \frac{1}{\ell} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = f^2$,$x = 1/\ell$,$m = \frac{g}{4 \pi^2}$,and $c = 0$.
Since the equation represents a linear relationship between $f^2$ and $1/\ell$,the graph between $f^2$ (on the ordinate) and $1/\ell$ (on the abscissa) will be a straight line passing through the origin.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
Let $L_1 = 121 \ cm = 1.21 \ m$ and $L_2 = 100 \ cm = 1.0 \ m$.
Let $n_1$ be the number of oscillations of the longer pendulum and $n_2$ be the number of oscillations of the shorter pendulum.
For them to be in phase at the mean position again,the total time must be equal: $n_1 T_1 = n_2 T_2$.
$n_1 (2 \pi \sqrt{\frac{1.21}{g}}) = n_2 (2 \pi \sqrt{\frac{1.0}{g}})$.
$n_1 (1.1) = n_2 (1.0)$.
$1.1 n_1 = n_2$,which implies $\frac{n_2}{n_1} = \frac{1.1}{1} = \frac{11}{10}$.
Since $n_2$ and $n_1$ must be integers,the minimum number of vibrations for the shorter pendulum $(n_2)$ is $11$ and for the longer pendulum $(n_1)$ is $10$.