SHM of Simple Pendulum Questions in English

(C) The time period $T$ of a simple pendulum is given by the formula:
$T = 2 \pi \sqrt{\frac{L}{g}}$
From this relation,we can see that $T \propto \sqrt{L}$.
If a pendulum clock is running slow,it means the time period $T$ is greater than the required value.
To correct the time,we need to decrease the time period $T$.
Since $T$ is directly proportional to the square root of the length $L$,decreasing the length $L$ will decrease the time period $T$.
Therefore,we should reduce the length of the pendulum.

(C) For a simple pendulum,the restoring torque is given by $\tau = -mgL \sin \theta$.
For small oscillations,$\sin \theta \approx \theta$,so $\tau \approx -mgL \theta$.
The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{-mgL \theta}{mL^2} = -\frac{g}{L} \theta$.
Comparing this with the $SHM$ equation $\alpha = -\omega^2 \theta$,we get $\omega^2 = \frac{g}{L}$.
The force constant $k$ for an oscillating system is defined as $k = m \omega^2$.
Substituting $\omega^2$,we get $k = m \left(\frac{g}{L}\right)$.
Thus,the force constant $k$ is directly proportional to the mass of the bob $m$ and inversely proportional to the length $L$ of the pendulum.
Among the given options,the most appropriate relationship is that it is directly proportional to the mass of the bob.

(A) The 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 trolley moves horizontally with acceleration $a$,the effective acceleration $g_{eff}$ is the vector sum of the gravitational acceleration $g$ (acting downwards) and the pseudo-acceleration $a$ (acting horizontally in the opposite direction).
Since these two accelerations are at right angles to each other,the magnitude of the effective acceleration is $g_{eff} = \sqrt{g^2 + a^2} = (a^2 + g^2)^{\frac{1}{2}}$.
Substituting this into the formula for the time period:
$T = 2 \pi \sqrt{\frac{L}{(a^2 + g^2)^{\frac{1}{2}}}} = 2 \pi \sqrt{L} \cdot (a^2 + g^2)^{-\frac{1}{4}}$.

(B) For a seconds pendulum,the time period $T = 2 \ s$.
The formula for the time period is $T = 2 \pi \sqrt{\frac{l}{g}}$.
Substituting $T = 2$,we get $2 = 2 \pi \sqrt{\frac{l}{g}}$,which simplifies to $1 = \pi \sqrt{\frac{l}{g}}$.
Squaring both sides,we get $1 = \pi^2 \frac{l}{g}$,so $l = \frac{g}{\pi^2}$.
At place $A$,$g_A = 981 \ cm/s^2$ and $\pi^2 = 10$,so $l_A = \frac{981}{10} = 98.1 \ cm$.
At place $B$,the length is decreased by $0.3 \ cm$,so $l_B = 98.1 - 0.3 = 97.8 \ cm$.
Since it is still a seconds pendulum,$T = 2 \ s$ at place $B$ as well.
Using $T = 2 \pi \sqrt{\frac{l_B}{g_B}}$,we have $1 = \pi^2 \frac{l_B}{g_B}$.
Therefore,$g_B = \pi^2 \times l_B = 10 \times 97.8 = 978 \ cm/s^2$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.
For a seconds pendulum, the time period $T = 2 \,s$ on both the earth and the planet.
Thus, $T_e = T_p = 2 \,s$.
This implies $\frac{\ell_e}{g_e} = \frac{\ell_p}{g_p}$, so $\ell_p = \ell_e \left( \frac{g_p}{g_e} \right)$.
The acceleration due to gravity is $g = \frac{GM}{R^2}$.
Given $M_p = 2M_e$ and $R_p = 2R_e$ (since diameter is double, radius is also double).
Therefore, $g_p = \frac{G(2M_e)}{(2R_e)^2} = \frac{2GM_e}{4R_e^2} = \frac{1}{2} g_e$.
Substituting this into the length equation: $\ell_p = 1 \,m \times \left( \frac{g_e/2}{g_e} \right) = 1 \times 0.5 = 0.5 \,m$.

(C) The given equation for displacement is $x = a \sin \left(\frac{\pi}{\sqrt{2}} t\right)$.
Comparing this with the standard $SHM$ equation $x = a \sin(\omega t)$,we get the angular frequency $\omega = \frac{\pi}{\sqrt{2}} \ rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/\sqrt{2}} = 2\sqrt{2} \ s$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g}}$.
Squaring both sides,we get $T^{2} = 4\pi^{2} \frac{\ell}{g}$.
Substituting $T = 2\sqrt{2}$ and $g = \pi^{2}$,we have $(2\sqrt{2})^{2} = 4\pi^{2} \frac{\ell}{\pi^{2}}$.
$8 = 4\ell$.
Therefore,$\ell = 2 \ m$.

(D) 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 relation, we can see that $T \propto \sqrt{L}$.
If we want to double the time period, let the new time period be $T' = 2T$.
Then, $2T = 2\pi \sqrt{\frac{L'}{g}}$.
Dividing the new equation by the original equation: $\frac{2T}{T} = \frac{2\pi \sqrt{L'/g}}{2\pi \sqrt{L/g}}$.
This simplifies to $2 = \sqrt{\frac{L'}{L}}$.
Squaring both sides, we get $4 = \frac{L'}{L}$, which implies $L' = 4L$.
Therefore, the length must be increased four times to double the time period.

(D) For a simple pendulum executing $S$.$H$.$M$. with small amplitude $A$,the tension $T$ in the string at any angle $\theta$ is given by $T = mg \cos \theta + \frac{mv^2}{\ell}$.
At the mean position,the velocity $v$ is maximum,and $\theta = 0$,so $\cos \theta = 1$. Thus,the maximum tension is $T_{\max} = mg + \frac{mv_{\max}^2}{\ell}$.
In $S$.$H$.$M$.,the velocity is given by $v = A\omega \cos(\omega t)$,where $\omega = \sqrt{\frac{g}{\ell}}$.
The maximum velocity is $v_{\max} = A\omega = A\sqrt{\frac{g}{\ell}}$.
Therefore,$v_{\max}^2 = A^2 \frac{g}{\ell}$.
Substituting this into the expression for $T_{\max}$:
$T_{\max} = mg + \frac{m}{\ell} \left( A^2 \frac{g}{\ell} \right) = mg + mg \frac{A^2}{\ell^2} = mg \left( 1 + \frac{A^2}{\ell^2} \right)$.

(A) The tension in the string is maximum when the bob passes through the mean position.
At the mean position,the forces acting on the bob are the tension $T$ upwards and weight $mg$ downwards. The net centripetal force is provided by the difference between tension and weight:
$T_{\max} - mg = \frac{mV^{2}}{L} \implies T_{\max} = mg + \frac{mV^{2}}{L} \dots (1)$
In Simple Harmonic Motion $(SHM)$,the velocity at the mean position is given by $V = a\omega$.
For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{L}}$.
Thus,$V = a\sqrt{\frac{g}{L}}$,which implies $V^{2} = a^{2}\frac{g}{L}$.
Substituting this value of $V^{2}$ into $Eq. (1)$:
$T_{\max} = mg + \frac{m}{L} \left(a^{2}\frac{g}{L}\right) = mg + \frac{mga^{2}}{L^{2}} = mg \left[1 + \left(\frac{a}{L}\right)^{2}\right]$.

(A) The total energy $E$ of a simple pendulum is given by the formula $E = \frac{1}{2} m \omega^2 A^2$, where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.
Since $\omega = \sqrt{\frac{g}{l}}$, the energy expression becomes $E = \frac{1}{2} m (\frac{g}{l}) A^2 = \frac{mgA^2}{2l}$.
Given that the amplitude $A$ remains constant and the length $l$ is changed to $l' = \frac{l}{3}$, the new energy $E'$ is $E' = \frac{mgA^2}{2(l/3)} = 3 \times \frac{mgA^2}{2l} = 3E$.
Therefore, the total energy becomes $3$ times the original energy.

(B) The total energy $E$ of a simple pendulum performing simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.
For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{L}}$,so $\omega^2 = \frac{g}{L}$.
Substituting this into the energy equation: $E = \frac{1}{2} m \left(\frac{g}{L}\right) A^2$.
Given $E_A = E_B$,$M_1 = M_2 = M$,and $L_1 = 2 L_2$:
$\frac{1}{2} M \left(\frac{g}{L_1}\right) A_A^2 = \frac{1}{2} M \left(\frac{g}{L_2}\right) A_B^2$.
Simplifying,we get $\frac{A_A^2}{L_1} = \frac{A_B^2}{L_2}$.
Substituting $L_1 = 2 L_2$: $\frac{A_A^2}{2 L_2} = \frac{A_B^2}{L_2}$.
This leads to $A_A^2 = 2 A_B^2$,or $A_A = \sqrt{2} A_B$.
Therefore,$A_B = \frac{A_A}{\sqrt{2}}$,which means the amplitude of $B$ is smaller than the amplitude of $A$.

(B) The total energy $(E)$ of a simple pendulum performing simple harmonic motion is given by the formula $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.
For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{l}}$,where $g$ is the acceleration due to gravity and $l$ is the length of the pendulum.
Substituting $\omega^2 = \frac{g}{l}$ into the energy equation,we get $E = \frac{1}{2} m (\frac{g}{l}) A^2$.
Since the mass $(m)$,acceleration due to gravity $(g)$,and amplitude $(A)$ remain constant,the total energy is inversely proportional to the length of the pendulum: $E \propto \frac{1}{l}$.
If the new length $l' = \frac{l}{4}$,then the new energy $E'$ is $E' \propto \frac{1}{l/4} = 4 \times (\frac{1}{l}) = 4E$.
Therefore,the total energy becomes $4$ times the initial energy.

(D) For a simple harmonic oscillator,the potential energy $PE$ at displacement $x$ is given by $PE = \frac{1}{2} k x^2$,where $k$ is the force constant.
At the extreme position,the displacement $x = A$,so $PE = \frac{1}{2} k A^2$.
For a simple pendulum,the restoring force is $F = -\frac{Mg}{L} x$,so the force constant $k = \frac{Mg}{L}$.
Substituting the value of $k$ into the potential energy formula:
$PE = \frac{1}{2} \left( \frac{Mg}{L} \right) A^2$.
$PE = \frac{MgA^2}{2L}$.

(D) The potential energy $(P.E.)$ of a simple harmonic oscillator at its extreme position is given by the formula:
$P.E. = \frac{1}{2} m \omega^2 A^2$
For a simple pendulum,the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{g}{L}}$.
Substituting this value into the potential energy equation:
$P.E. = \frac{1}{2} m \left(\sqrt{\frac{g}{L}}\right)^2 A^2$
$P.E. = \frac{1}{2} m \left(\frac{g}{L}\right) A^2$
$P.E. = \frac{mgA^2}{2L}$

(C) The swing performs $S.H.M$ between the minimum height $(h_{min} = 0.75 \,m)$ and the maximum height $(h_{max} = 2 \,m)$.
At the highest point,the potential energy is maximum,and at the lowest point,this potential energy is converted into kinetic energy.
The effective vertical displacement (amplitude height) is $h = h_{max} - h_{min} = 2 \,m - 0.75 \,m = 1.25 \,m$.
Using the principle of conservation of energy:
$\frac{1}{2} mv^2 = mgh$
$v^2 = 2gh$
$v^2 = 2 \times 10 \,m/s^2 \times 1.25 \,m$
$v^2 = 25 \,m^2/s^2$
$v = 5 \,m/s$.

(A) For a simple pendulum undergoing small oscillations,the restoring force is $F = -\frac{mg}{L} x$.
The potential energy $PE$ of a simple harmonic oscillator is given by $PE = \frac{1}{2} k x^{2}$.
Comparing the restoring force $F = -kx$ with $F = -\frac{mg}{L} x$,we get the spring constant $k = \frac{mg}{L}$.
At the extreme position,the displacement $x$ is equal to the amplitude $A$.
Substituting these values into the potential energy formula: $PE = \frac{1}{2} \left( \frac{mg}{L} \right) A^{2} = \frac{mgA^{2}}{2L}$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$.
Since $\omega = \frac{2 \pi}{T}$,we have $\omega = \sqrt{\frac{g}{\ell}}$,which implies $\omega \propto \frac{1}{\sqrt{\ell}}$.
Given the new length $\ell_2 = \frac{\ell_1}{3}$,the new angular frequency $\omega_2$ is related to the initial frequency $\omega_1$ by $\frac{\omega_2}{\omega_1} = \sqrt{\frac{\ell_1}{\ell_2}} = \sqrt{\frac{\ell_1}{\ell_1/3}} = \sqrt{3}$.
The total energy of a simple harmonic oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$.
Since the mass $(m)$ and amplitude $(A)$ remain constant,$E \propto \omega^2$.
Therefore,$\frac{E_2}{E_1} = \left( \frac{\omega_2}{\omega_1} \right)^2 = (\sqrt{3})^2 = 3$.
Thus,the new total energy is $E_2 = 3 E_1$.

(B) The tension $T$ in the string of a simple pendulum at any angle $\theta$ is given by the radial force balance equation:
$T = mg \cos \theta + \frac{mv^2}{l}$
For small oscillations,$\cos \theta \approx 1$ and the velocity $v$ is maximum at the mean position $(\theta = 0)$.
Thus,the maximum tension $T_{\max}$ occurs at the mean position:
$T_{\max} = mg + \frac{mv_{\max}^2}{l}$
For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{l}}$.
The maximum velocity in $S$.$H$.$M$. is $v_{\max} = A\omega$.
Substituting $v_{\max}$ into the tension equation:
$T_{\max} = mg + \frac{m(A\omega)^2}{l}$
$T_{\max} = mg + \frac{m A^2 \omega^2}{l}$
Since $\omega^2 = \frac{g}{l}$,we get:
$T_{\max} = mg + \frac{m A^2 (g/l)}{l}$
$T_{\max} = mg + \frac{m A^2 g}{l^2}$
$T_{\max} = mg \left[1 + \left(\frac{A}{l}\right)^2\right]$

(B) The path length of oscillation is the total distance between the two extreme positions, which is equal to $2a$, where $a$ is the amplitude.
Given, $2a = 16 \,cm$, so the amplitude $a = 8 \,cm$.
The length of the pendulum is $l = 1 \,m$.
The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{l}}$.
Substituting $g = \pi^2 \,m/s^2$ and $l = 1 \,m$, we get $\omega = \sqrt{\frac{\pi^2}{1}} = \pi \,rad/s$.
The maximum velocity $v_{max}$ in simple harmonic motion is given by $v_{max} = a\omega$.
Substituting the values, $v_{max} = 8 \,cm \times \pi \,rad/s = 8\pi \,cm/s$.

(D) Given: Length $L = 0.4 \ m$,maximum speed $v_{max} = 4 \ m/s$ at the lowest point.
When the string makes an angle of $30^{\circ}$ with the horizontal,it makes an angle of $\theta = 60^{\circ}$ with the vertical.
The height $h$ of the pendulum bob above the lowest point is given by $h = L - L \cos \theta = L(1 - \cos 60^{\circ})$.
$h = 0.4(1 - 0.5) = 0.4 \times 0.5 = 0.2 \ m$.
Using the law of conservation of energy between the lowest point and the point at angle $\theta$:
$\frac{1}{2} m v_{max}^{2} = \frac{1}{2} m v^{2} + mgh$
$v^{2} = v_{max}^{2} - 2gh$
$v^{2} = (4)^{2} - 2 \times 10 \times 0.2$
$v^{2} = 16 - 4 = 12$
$v = \sqrt{12} = 2 \sqrt{3} \ m/s$.

(A) The coin rests on the plate and moves with it. The acceleration of the plate in $S.H.M.$ is given by $a = \omega^{2} x$,where $x$ is the displacement from the mean position.
As the plate moves downward,its acceleration is directed upwards. As the plate moves upward,its acceleration is directed downwards.
The coin loses contact with the plate when the downward acceleration of the plate exceeds the acceleration due to gravity $(g)$.
At the topmost point of the oscillation,the downward acceleration is maximum,given by $a_{max} = \omega^{2} A$.
For the coin to lose contact,the condition is $a_{max} \geq g$.
Therefore,the minimum amplitude at which contact is lost is $A = \frac{g}{\omega^{2}}$.

(A) The potential energy of a simple harmonic oscillator is given by the formula $U = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
For a simple pendulum,the restoring force is $F = -Mg \sin \theta \approx -Mg \theta$ for small angles.
Since $\theta = \frac{x}{L}$,the force is $F = -\frac{Mg}{L} x$.
Comparing this with $F = -kx$,we get the force constant $k = \frac{Mg}{L}$.
Substituting $k$ into the potential energy formula: $U = \frac{1}{2} (\frac{Mg}{L}) A^2 = \frac{MgA^2}{2L}$.

(C) The frequency $f$ of a simple pendulum is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$,where $g$ is the acceleration due to gravity and $L$ is the length of the pendulum.
Since the pendulums are at the same place,$g$ is constant.
Therefore,$f \propto \frac{1}{\sqrt{L}}$,which implies $f^2 \propto \frac{1}{L}$ or $L \propto \frac{1}{f^2}$.
Given the ratio of frequencies $f_1 : f_2 = 4 : 3$,we have $\frac{f_1}{f_2} = \frac{4}{3}$.
The ratio of their lengths is $\frac{L_1}{L_2} = \frac{f_2^2}{f_1^2} = \left( \frac{f_2}{f_1} \right)^2$.
Substituting the values,$\frac{L_1}{L_2} = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.
Thus,the ratio of their lengths is $9: 16$.

(A) The time period of a simple pendulum is given by the formula $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$,where $l$ is the length of the pendulum and $g_{eff}$ is the effective acceleration due to gravity.
When the point of suspension moves vertically upward with an acceleration '$a$',the effective acceleration due to gravity becomes $g_{eff} = g + a$.
Substituting this into the formula,we get $T' = 2 \pi \sqrt{\frac{l}{g + a}}$.
Since $g + a > g$,the denominator increases,which causes the time period $T'$ to decrease compared to the original time period $T$.

(B) We know that the time period $T$ of a simple pendulum is given by the formula:
$T = 2 \pi \sqrt{\frac{l}{g}}$
From this relation,we can see that $T \propto \sqrt{l}$.
Let the initial time period be $T_1$ with length $l_1$,and the final time period be $T_2$ with length $l_2$.
We want the period to be doubled,so $T_2 = 2 T_1$.
Using the ratio:
$\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$
Substituting $T_2 = 2 T_1$:
$2 = \sqrt{\frac{l_2}{l_1}}$
Squaring both sides:
$4 = \frac{l_2}{l_1} \Rightarrow l_2 = 4 l_1$
Therefore,the period of a simple pendulum is doubled when its length is made four times the original length.

(C) For a simple pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Thus,$T_1 = 2 \pi \sqrt{\frac{l_1}{g}}$ and $T_2 = 2 \pi \sqrt{\frac{l_2}{g}}$.
Squaring both sides,we get $T_1^2 = 4 \pi^2 \frac{l_1}{g}$ and $T_2^2 = 4 \pi^2 \frac{l_2}{g}$.
From these,we can express the lengths as $l_1 = \frac{T_1^2 g}{4 \pi^2}$ and $l_2 = \frac{T_2^2 g}{4 \pi^2}$.
Now,for a pendulum of length $(l_1 - l_2)$,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{l_1 - l_2}{g}}$.
Substituting the expressions for $l_1$ and $l_2$:
$T = 2 \pi \sqrt{\frac{1}{g} \left( \frac{T_1^2 g}{4 \pi^2} - \frac{T_2^2 g}{4 \pi^2} \right)}$.
$T = 2 \pi \sqrt{\frac{g}{g \cdot 4 \pi^2} (T_1^2 - T_2^2)}$.
$T = 2 \pi \sqrt{\frac{1}{4 \pi^2} (T_1^2 - T_2^2)}$.
$T = 2 \pi \cdot \frac{1}{2 \pi} \sqrt{T_1^2 - T_2^2}$.
$T = \sqrt{T_1^2 - T_2^2}$.

(C) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.
From this formula,we can see that the time period $T$ is directly proportional to the square root of the length $l$,i.e.,$T \propto \sqrt{l}$.
The time period of a simple pendulum is independent of the mass,size,or density of the bob.
Let the initial length be $l_1 = l$ and the initial time period be $T_1 = T$.
Let the new length be $l_2$ and the new time period be $T_2 = 2T$.
Using the proportionality $T \propto \sqrt{l}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$.
Substituting the given values: $\frac{2T}{T} = \sqrt{\frac{l_2}{l}}$.
$2 = \sqrt{\frac{l_2}{l}}$.
Squaring both sides,we get $4 = \frac{l_2}{l}$.
Therefore,the new length $l_2 = 4l$.

(D) The time period of $S.H.M.$ for small vertical oscillations of a floating body is given by $T = 2 \pi \sqrt{\frac{h'}{g}}$,where $h'$ is the depth of the object submerged in the liquid.
For a floating body,the weight of the object equals the weight of the displaced liquid.
Let $V$ be the volume of the wood,$V = a \times b \times c$.
Mass of wood $= V \times d \times \rho_w = (abc) \times d \times \rho_w$ (where $\rho_w$ is the density of water).
Volume of displaced water $V_{disp} = b \times c \times h'$.
Weight of displaced water $= (bc h') \times \rho_w \times g$.
Equating the two: $(abc) \times d \times \rho_w \times g = (bc h') \times \rho_w \times g$.
Solving for $h'$,we get $h' = ad$.
Substituting this into the time period formula: $T = 2 \pi \sqrt{\frac{ad}{g}}$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,where $L$ is the distance between the point of suspension and the center of gravity $(CG)$ of the bob.
Initially,the $CG$ of the water-filled ball is at its geometric center.
As water flows out through the small hole,the $CG$ of the remaining water shifts downwards,which increases the effective length $L$ of the pendulum,causing the time period $T$ to increase.
As the ball becomes nearly empty,the $CG$ starts moving back up towards the geometric center of the ball.
Consequently,the effective length $L$ decreases and returns to its original value.
Therefore,the time period first increases and then decreases,finally returning to its initial value.

(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$.
In a stationary lift,the effective acceleration is $g_{eff} = g$,so $T = 2 \pi \sqrt{\frac{l}{g}}$.
When the lift accelerates upwards with an acceleration $a = \frac{g}{3}$,the effective acceleration experienced by the pendulum is $g_{eff} = g + a$.
Substituting the value of $a$,we get $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$.
The new time period $T^{\prime}$ is given by $T^{\prime} = 2 \pi \sqrt{\frac{l}{g_{eff}}} = 2 \pi \sqrt{\frac{l}{4g/3}}$.
Simplifying this,$T^{\prime} = 2 \pi \sqrt{\frac{3l}{4g}} = \frac{\sqrt{3}}{2} \left( 2 \pi \sqrt{\frac{l}{g}} \right)$.
Since $T = 2 \pi \sqrt{\frac{l}{g}}$,we have $T^{\prime} = \frac{\sqrt{3}}{2} T$.

(C) The time period $T$ of a simple pendulum is given by the formula:
$T = 2\pi \sqrt{\frac{\ell}{g}}$
From this relation,we can see that $T \propto \sqrt{\ell}$.
If a clock is running fast,it means its time period $T$ is too small (it completes oscillations too quickly).
To correct the time,we need to increase the time period $T$.
Since $T$ is directly proportional to the square root of the length $\ell$,increasing the length $\ell$ will increase the time period $T$.
The time period is independent of the mass of the bob and the amplitude of oscillation (for small angles).

(A) The frequency $f$ of a simple pendulum is given by the formula $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$.
From this,we can see that $f \propto \frac{1}{\sqrt{\ell}}$,which implies $\frac{f_1}{f_2} = \sqrt{\frac{\ell_2}{\ell_1}}$.
Given the ratio of frequencies $\frac{f_1}{f_2} = \frac{3}{4}$,we substitute this into the equation:
$\frac{3}{4} = \sqrt{\frac{\ell_2}{\ell_1}}$.
Squaring both sides,we get $\frac{9}{16} = \frac{\ell_2}{\ell_1}$.
Therefore,the ratio of their lengths is $\frac{\ell_1}{\ell_2} = \frac{16}{9}$.

(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{\ell}{g}} = \sqrt{3} \ s$.
When the lift moves upward with an acceleration $a = \frac{g}{3}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{3} = \frac{4g}{3}$.
The new time period $T^{\prime}$ is given by $T^{\prime} = 2\pi \sqrt{\frac{\ell}{g_{eff}}} = 2\pi \sqrt{\frac{\ell}{4g/3}}$.
Dividing $T^{\prime}$ by $T$,we get $\frac{T^{\prime}}{T} = \sqrt{\frac{g}{4g/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Substituting the value of $T = \sqrt{3} \ s$,we get $T^{\prime} = \sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2} = 1.5 \ s$.

(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the string and $g$ is the acceleration due to gravity.
When the lift accelerates upwards with an acceleration $a = \frac{g}{3}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{3} = \frac{4g}{3}$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g_{eff}}} = 2\pi \sqrt{\frac{l}{4g/3}} = 2\pi \sqrt{\frac{3l}{4g}}$.
This can be rewritten as $T' = \frac{\sqrt{3}}{2} \times 2\pi \sqrt{\frac{l}{g}}$.
Substituting $T = 2\pi \sqrt{\frac{l}{g}}$,we get $T' = \frac{\sqrt{3}}{2} T$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
In air,the effective acceleration due to gravity is $g_{eff} = g$.
In water,the bob experiences an upward buoyant force. The effective acceleration is $g_{eff}' = g(1 - \frac{\rho}{\sigma})$,where $\rho$ is the density of water $(10^3 \ kg/m^3)$ and $\sigma$ is the density of the bob $(\frac{9}{8} \times 10^3 \ kg/m^3)$.
$g_{eff}' = g(1 - \frac{10^3}{\frac{9}{8} \times 10^3}) = g(1 - \frac{8}{9}) = g(\frac{1}{9})$.
Now,the time period in water is $T_1 = 2\pi \sqrt{\frac{l}{g_{eff}'}} = 2\pi \sqrt{\frac{l}{g/9}} = 2\pi \sqrt{\frac{9l}{g}}$.
$T_1 = 3 \times (2\pi \sqrt{\frac{l}{g}}) = 3T$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{L/g}$,where $L$ is the distance between the point of suspension and the center of mass $(CM)$ of the system.
Initially,the $CM$ of the water-filled sphere is at the geometric center of the sphere.
As water flows out from the bottom hole,the $CM$ of the remaining water shifts downwards,which increases the effective length $L$ of the pendulum,causing the time period $T$ to increase.
As the sphere becomes nearly empty,the $CM$ of the remaining water shifts back upwards towards the center of the sphere.
This causes the effective length $L$ to decrease,which in turn causes the time period $T$ to decrease.
Therefore,the period of oscillation first increases and then decreases.

(D) The time period of a simple pendulum is given by the formula:
$T = 2 \pi \sqrt{\frac{l}{g}}$
Here,$l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
Squaring both sides of the equation,we get:
$T^2 = \frac{4 \pi^2 l}{g}$
This can be written as:
$T^2 = k \cdot l$,where $k = \frac{4 \pi^2}{g}$ is a constant.
This equation is of the form $y^2 = 4ax$,which represents a parabola.
Therefore,the graph between the time period $T$ and the length $l$ is a part of a parabola.

(C) Given: Length of pendulum $l = 2 \ m$,angular displacement $\theta = 60^{\circ}$,mass $m = 200 \ g = 0.2 \ kg$,acceleration due to gravity $g = 10 \ m/s^2$.
When the pendulum bob moves in a horizontal circle,the radius of the circular path is $r = l \sin \theta$.
For the bob to move in a horizontal circle,the forces acting on it are tension $T$ and weight $mg$.
The vertical component of tension balances the weight: $T \cos \theta = mg$ ... $(i)$
The horizontal component of tension provides the necessary centripetal force: $T \sin \theta = m r \omega^2 = m (l \sin \theta) \omega^2$ ... (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{T \sin \theta}{T \cos \theta} = \frac{m l \sin \theta \omega^2}{mg}$
$\tan \theta = \frac{l \omega^2 \sin \theta}{g}$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$\frac{1}{\cos \theta} = \frac{l \omega^2}{g}$
$\omega^2 = \frac{g}{l \cos \theta}$
Substituting the values: $\omega^2 = \frac{10}{2 \times \cos 60^{\circ}} = \frac{10}{2 \times 0.5} = \frac{10}{1} = 10$
$\omega = \sqrt{10} = \sqrt{4 \times 2.5} = 2 \sqrt{2.5} \ rad/s$.

(B) Let $m$ be the mass of the bob and $l$ be the length of the string. The tension in the string at any angle $\phi$ is given by $T = mg \cos \phi + \frac{mv^2}{l}$.
At the extreme position,$\phi = \theta$ and $v = 0$,so the minimum tension is $T_{min} = mg \cos \theta$.
At the lowest point,$\phi = 0$. Using conservation of energy,$mg(l - l \cos \theta) = \frac{1}{2}mv^2$,so $mv^2 = 2mgl(1 - \cos \theta)$.
The maximum tension is $T_{max} = mg + \frac{mv^2}{l} = mg + 2mg(1 - \cos \theta) = mg(3 - 2 \cos \theta)$.
Given $T_{max} = 4 T_{min}$,we have $mg(3 - 2 \cos \theta) = 4mg \cos \theta$.
$3 - 2 \cos \theta = 4 \cos \theta \implies 6 \cos \theta = 3 \implies \cos \theta = 0.5$.
Therefore,$\theta = \cos^{-1}(0.5)$.

(A) simple pendulum,in practice,consists of a heavy but small-sized metallic bob suspended by a light,inextensible,and flexible string.
It oscillates simple harmonically if and only if:
$(I)$ The size of the bob is negligible compared to the length of the string of the pendulum,which allows us to treat the bob as a point mass.
$(II)$ The angular amplitude (the angle between the vertical mean position and the string at the extreme point) is small,typically less than $10^{\circ}$,so that the approximation $\sin \theta \approx \theta$ holds true.
Thus,both statements are correct.

(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Given $l = 1 \,m$ and taking $g = 10 \,m/s^2$ (or $\pi^2 \approx 10$),we have $T = 2 \pi \sqrt{\frac{1}{10}} \approx 2 \,s$.
For a simple pendulum,the time taken to travel from the extreme position to the mean position $(B)$ is $\frac{T}{4}$.
Since both spheres are released from their respective extreme positions,they will both reach the mean position $B$ at time $t = \frac{T}{4} = \frac{2}{4} = 0.5 \,s$.
Therefore,the two spheres will collide at the mean position $B$ after $0.5 \,s$.

(B) The time period of a simple pendulum at rest is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the lift accelerates upwards with an acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = g + a$.
The new time period is $T' = 2\pi \sqrt{\frac{l}{g+a}}$.
Given that $T' = T/2$,we have $\frac{T}{2} = 2\pi \sqrt{\frac{l}{g+a}}$.
Dividing the expression for $T'$ by $T$,we get $\frac{1}{2} = \sqrt{\frac{g}{g+a}}$.
Squaring both sides,$\frac{1}{4} = \frac{g}{g+a}$.
This implies $g + a = 4g$,so $a = 3g$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
For pendulum $A$: It completes $10$ oscillations in $20 \ s$.
Therefore,the time period $T_A = \frac{20}{10} = 2 \ s$.
$2\pi \sqrt{\frac{l_A}{g}} = 2 \quad (1)$
For pendulum $B$: It completes $8$ oscillations in $10 \ s$.
Therefore,the time period $T_B = \frac{10}{8} = \frac{5}{4} \ s$.
$2\pi \sqrt{\frac{l_B}{g}} = \frac{5}{4} \quad (2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{2\pi \sqrt{l_A/g}}{2\pi \sqrt{l_B/g}} = \frac{2}{5/4}$
$\sqrt{\frac{l_A}{l_B}} = \frac{2 \times 4}{5} = \frac{8}{5}$
Squaring both sides:
$\frac{l_A}{l_B} = \left(\frac{8}{5}\right)^2 = \frac{64}{25}$

(A) Given: Mass $m = 40 \ g = 0.04 \ kg$,charge $q = 2 \times 10^{-6} \ C$,$g = 10 \ ms^{-2}$,electric field $E = 4.2 \times 10^4 \ NC^{-1}$.
First,calculate the acceleration due to the electric field:
$a = \frac{qE}{m} = \frac{2 \times 10^{-6} \times 4.2 \times 10^4}{0.04} = \frac{0.084}{0.04} = 2.1 \ ms^{-2}$.
Since the electric field is downward and the charge is positive,the force is downward,so the effective acceleration $g_e = g + a = 10 + 2.1 = 12.1 \ ms^{-2}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
In the absence of the field,$T = \frac{44}{20} = 2.2 \ s$.
Thus,$2.2 = 2\pi \sqrt{\frac{l}{10}} \Rightarrow \sqrt{l} = \frac{2.2 \sqrt{10}}{2\pi}$.
In the presence of the field,$T' = 2\pi \sqrt{\frac{l}{12.1}} = 2\pi \frac{\sqrt{l}}{\sqrt{12.1}}$.
Substituting $\sqrt{l}$,we get $T' = 2\pi \left( \frac{2.2 \sqrt{10}}{2\pi \sqrt{12.1}} \right) = 2.2 \sqrt{\frac{10}{12.1}} = 2.2 \times \frac{\sqrt{10}}{1.1 \sqrt{10}} = 2.2 \times \frac{1}{1.1} = 2 \ s$.
The time taken for $15$ oscillations is $t = 15 \times T' = 15 \times 2 = 30 \ s$.

(B) Let the mass of the pendulum bob be $m$ and the length of the string be $l$. The center of mass of the string is at $l/2$ from the pivot.
When the pendulum is displaced by $90^{\circ}$ and released,the change in potential energy is converted into kinetic energy.
Using the law of conservation of energy: $m g (l/2) = (1/2) m v^2$.
This gives $v^2 = g l$.
At the mean position,the tension $T$ in the string must balance the weight $m g$ and provide the necessary centripetal force $F_c = m v^2 / r$,where $r = l/2$ is the distance of the center of mass from the pivot.
Therefore,$T = m g + (m v^2) / (l/2)$.
Substituting $v^2 = g l$ into the equation:
$T = m g + (2 m / l) \cdot (g l) = m g + 2 m g = 3 m g$.
Thus,the minimum strength of the string must be $3 m g$.

(D) When the object is released from rest, its velocity is zero. Therefore, the centripetal acceleration $(a_c = v^2/r)$ is zero.
The forces acting on the object are its weight $(mg)$ acting vertically downwards and the tension $(T)$ in the string acting along the string.
We resolve the weight $(mg)$ into two components: one along the string $(mg \cos 30^{\circ})$ and one perpendicular to the string $(mg \sin 30^{\circ})$.
Since the string remains taut and the object starts from rest, the net force along the string must be zero to satisfy the condition of no radial motion at the instant of release $(T - mg \cos 30^{\circ} = 0)$.
The net force acting on the object is the component perpendicular to the string, which is $F_{net} = mg \sin 30^{\circ}$.
Using Newton's second law, $F_{net} = ma$, we get:
$ma = mg \sin 30^{\circ}$
$a = g \sin 30^{\circ}$
Given $g = 10 \,ms^{-2}$ and $\sin 30^{\circ} = 0.5$:
$a = 10 \times 0.5 = 5.0 \,ms^{-2}$.

(A) The formula for the time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.
Given: $T = 4 \ s$ and $g = \pi^2 \ ms^{-2}$.
Substituting these values into the formula:
$4 = 2\pi \sqrt{\frac{\ell}{\pi^2}}$
Squaring both sides:
$16 = 4\pi^2 \cdot \frac{\ell}{\pi^2}$
$16 = 4\ell$
$\ell = \frac{16}{4} = 4 \ m$.
Therefore,the length of the pendulum is $4 \ m$.

(C) The frequency of a simple pendulum is given by the formula:
$f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$
From this formula, we can see that the frequency is inversely proportional to the square root of the length of the pendulum:
$f \propto \frac{1}{\sqrt{\ell}}$
Let the initial length be $\ell_1 = \ell$ and the initial frequency be $f_1 = 8 \,Hz$.
When the string is clamped at its midpoint, the new length becomes $\ell_2 = \frac{\ell}{2}$.
Now, we can find the ratio of the new frequency $f_2$ to the initial frequency $f_1$:
$\frac{f_2}{f_1} = \sqrt{\frac{\ell_1}{\ell_2}} = \sqrt{\frac{\ell}{\ell / 2}} = \sqrt{2}$
Therefore, the new frequency $f_2$ is:
$f_2 = \sqrt{2} \times f_1 = \sqrt{2} \times 8 \,Hz \approx 1.414 \times 8 \,Hz = 11.312 \,Hz$.
Rounding to the nearest provided option, we get $11.28 \,Hz$.

(A) simple pendulum consists of a point mass (bob) suspended by a massless,inextensible string.
For a simple pendulum oscillating in a plane,the position of the bob is uniquely determined by a single coordinate,which is the angle $\theta$ that the string makes with the vertical.
Therefore,the number of degrees of freedom for an oscillating simple pendulum is $1$.