SHM of Simple Pendulum Questions in English

(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g_{\text{eff}}}}$.
$(a)$ When the lift is stationary $(a = 0)$,the effective acceleration is $g_{\text{eff}} = g$. Thus,$T = 2 \pi \sqrt{\frac{\ell}{g}}$.
$(b)$ When the lift accelerates upwards with $a = \frac{g}{6}$,the effective acceleration becomes $g_{\text{eff}} = g + a = g + \frac{g}{6} = \frac{7g}{6}$.
The new time period $T^{\prime}$ is given by:
$T^{\prime} = 2 \pi \sqrt{\frac{\ell}{g_{\text{eff}}}} = 2 \pi \sqrt{\frac{\ell}{7g/6}} = 2 \pi \sqrt{\frac{6\ell}{7g}}$.
Comparing this with the original time period $T$,we get:
$T^{\prime} = \sqrt{\frac{6}{7}} \left( 2 \pi \sqrt{\frac{\ell}{g}} \right) = \sqrt{\frac{6}{7}} T$.

(A) The maximum velocity of the pendulum bob is attained at its lowest position due to the conservation of mechanical energy.
Let the length of the string be $\ell = 250\,cm = 2.5\,m$.
The vertical height $h$ through which the bob falls is given by:
$h = \ell - \ell \cos 60^{\circ} = \ell(1 - \cos 60^{\circ})$
$h = 2.5 \times (1 - 0.5) = 2.5 \times 0.5 = 1.25\,m$
Using the conservation of energy,the potential energy at the highest point is converted into kinetic energy at the lowest point:
$mgh = \frac{1}{2}mv_{\max}^2$
$v_{\max} = \sqrt{2gh}$
$v_{\max} = \sqrt{2 \times 10 \times 1.25} = \sqrt{25} = 5\,m/s$

(C) The standard equation for $S.H.M.$ is $Y = A \sin(\omega t + \phi)$.
Comparing this with the given equation $Y = A \sin(\pi t + \phi)$,we get the angular frequency $\omega = \pi \, rad/s$.
For a simple pendulum,the angular frequency is given by $\omega = \sqrt{\frac{g}{\ell}}$.
Squaring both sides,we get $\omega^2 = \frac{g}{\ell}$,which implies $\ell = \frac{g}{\omega^2}$.
Using the value of acceleration due to gravity $g = 980 \, cm/s^2$ and $\omega = \pi \, rad/s$:
$\ell = \frac{980}{\pi^2} \approx \frac{980}{9.8696} \approx 99.3 \, cm$.
Rounding to the nearest option,we get $\ell \approx 99.4 \, cm$.

(B) The effective acceleration due to gravity $g'$ when the bob is immersed in water is given by:
$g' = g - \frac{F_B}{m} = g - \frac{\rho_w V g}{\rho_b V} = g \left(1 - \frac{\rho_w}{\rho_b}\right)$
Given relative density of the bob $\rho_b / \rho_w = 5$,so $\rho_w / \rho_b = 1/5$.
$g' = g \left(1 - \frac{1}{5}\right) = g \left(\frac{4}{5}\right) = 0.8g$
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.
Therefore,the new time period $T'$ is:
$T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{0.8g}} = T \sqrt{\frac{1}{0.8}} = T \sqrt{\frac{5}{4}}$
Given $T = 10 \, s$,we have:
$T' = 10 \sqrt{\frac{5}{4}} = 10 \frac{\sqrt{5}}{2} = 5 \sqrt{5} \, s$
Comparing this with $5 \sqrt{x} \, s$,we get $x = 5$.

(A) When a vehicle moves down a frictionless inclined plane of inclination $\alpha$,its acceleration is $a = g \sin \alpha$ directed down the plane.
To find the effective acceleration due to gravity $(g_{eff})$ inside the vehicle,we consider the pseudo-force acting on the pendulum bob.
The effective acceleration is the vector sum of the acceleration due to gravity $(g)$ and the negative of the vehicle's acceleration $(-a)$.
$g_{eff} = \vec{g} - \vec{a}$.
Resolving these into components perpendicular to the inclined plane,we find $g_{eff} = g \cos \alpha$.
The time period of a simple pendulum is given by $T = 2 \pi \sqrt{L / g_{eff}}$.
Substituting $g_{eff} = g \cos \alpha$,we get $T = 2 \pi \sqrt{\frac{L}{g \cos \alpha}}$.

(A) The time period $T$ of a simple pendulum for large amplitudes is given by the relation:
$T = 2\pi \sqrt{\frac{l}{g}} \left( 1 + \frac{\theta_{0}^{2}}{16} \right)$
where $\theta_{0}$ is the angular amplitude from the mean position.
As the amplitude $\theta_{0}$ increases,the time period $T$ also increases.
The relationship between $T$ and $\theta_{0}$ is parabolic,meaning $T$ increases with $\theta_{0}$ in a non-linear,upward-curving fashion.
Therefore,the correct plot is the one showing $T$ increasing with $\theta_{0}$,which corresponds to Plot $A$.

(C) When the balloon is slightly displaced horizontally,the horizontal component of the buoyant force produces a torque about the end of the string attached to the ground. This torque produces sideways oscillations of the balloon.
The restoring torque is given by:
$\tau_1 = F_b \sin \theta \times l = V(\rho_{air} - \rho_{He})g l \sin \theta$
For small angular displacements,$\sin \theta \approx \theta$,so:
$\tau_1 = V(\rho_{air} - \rho_{He})g l \theta$
The inertial torque on the balloon is:
$\tau_2 = I \alpha = (m l^2) \alpha = (V \rho_{He}) l^2 \alpha$
Equating the torques (considering the restoring nature):
$V \rho_{He} l^2 \alpha = -V(\rho_{air} - \rho_{He}) g l \theta$
$\alpha = -\left(\frac{\rho_{air} - \rho_{He}}{\rho_{He}}\right) \frac{g}{l} \theta$
Comparing with the standard $SHM$ equation $\alpha = -\omega^2 \theta$,we get:
$\omega^2 = \left(\frac{\rho_{air} - \rho_{He}}{\rho_{He}}\right) \frac{g}{l}$
The time period $T$ is:
$T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\left(\frac{\rho_{He}}{\rho_{air} - \rho_{He}}\right) \frac{l}{g}}$

(C) For a simple pendulum,the equation of motion is given by $\frac{d^2 \theta}{dt^2} + \frac{g}{l} \sin \theta = 0$.
For small oscillations,$\sin \theta \approx \theta$,which leads to the simple harmonic motion period $T_0 = 2 \pi \sqrt{\frac{l}{g}}$.
However,for larger amplitudes (like $45^{\circ}$),we use the expansion $\sin \theta \approx \theta - \frac{\theta^3}{6}$.
The time period $T$ for a pendulum with amplitude $\theta_0$ is given by the formula $T = T_0 \left( 1 + \frac{\theta_0^2}{16} + \dots \right)$ (where $\theta_0$ is in radians).
Since $\theta_0 = 45^{\circ} = \frac{\pi}{4} \text{ radians}$,the term $\frac{\theta_0^2}{16}$ is positive.
Therefore,the time period $T$ will be slightly more than $T_0$.

(B) Let $v$ be the velocity of the bob at position $\theta$ after being released from the horizontal position.
From the law of conservation of energy,the potential energy lost equals the kinetic energy gained:
$mgh = \frac{1}{2}mv^2$
Since the height $h$ dropped from the horizontal is $l \cos \theta$,we have:
$mg(l \cos \theta) = \frac{1}{2}mv^2$
$\Rightarrow \frac{v^2}{l} = 2g \cos \theta$
The radial (centripetal) acceleration is $a_c = \frac{v^2}{l} = 2g \cos \theta$.
The tangential acceleration is due to the component of gravity perpendicular to the string:
$a_t = g \sin \theta$.
If the total acceleration vector $\vec{a}$ makes an angle $\phi$ with the string (which is the direction of the radial acceleration),then:
$\tan \phi = \frac{a_t}{a_c}$
Substituting the expressions for $a_t$ and $a_c$:
$\tan \phi = \frac{g \sin \theta}{2g \cos \theta} = \frac{\tan \theta}{2}$
Therefore,$\phi = \tan^{-1}\left(\frac{\tan \theta}{2}\right)$.

(A) For a simple pendulum undergoing simple harmonic motion,the displacement $x$ is given by $x = A \sin(\omega t)$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The tangential acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t)$.
From these equations,we have:
$\frac{v}{A\omega} = \cos(\omega t)$ and $\frac{a}{-A\omega^2} = \sin(\omega t)$.
Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$,we get:
$\left(\frac{v}{A\omega}\right)^2 + \left(\frac{a}{-A\omega^2}\right)^2 = 1$
$\frac{v^2}{A^2\omega^2} + \frac{a^2}{A^2\omega^4} = 1$
This is the equation of an ellipse in the $v-a$ plane,where $v$ is on the horizontal axis and $a$ is on the vertical axis. Thus,the correct graph is $I$.

(A) The correct option is $A$.
In a simple pendulum,the total mechanical energy is conserved in the absence of friction. The total energy of the pendulum at its extreme positions is purely potential energy,given by $U = mgh$,where $h$ is the vertical height of the bob from the lowest point of the oscillation.
Since the pendulum starts from point $A$ with zero initial velocity,its total energy is $E = mgh_A$. At the other extreme point $D$,the velocity is again zero,so its total energy is $E = mgh_D$.
By the law of conservation of energy,$mgh_A = mgh_D$,which implies $h_A = h_D$. Since $A$ and $B$ lie on the same horizontal line,$h_A = h_B$. Therefore,$h_D = h_B$,which means the extreme point $D$ must lie on the same horizontal line $A B$.

(B) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l_{eff}}{g}}$,where $l_{eff}$ is the effective length of the pendulum.
$l_{eff}$ is the distance between the point of suspension and the center of mass of the bob.
Initially,when the bob is full of water,the center of mass is at the geometric center of the bob.
As water flows out,the center of mass of the system (bob + remaining water) shifts downwards,which increases the effective length $l_{eff}$. Consequently,the time period $T$ increases.
As the water level continues to drop and the bob becomes empty,the center of mass eventually rises back to the geometric center of the bob.
Since the effective length $l_{eff}$ decreases during this second phase,the time period $T$ decreases.
Therefore,the time period $T$ first increases and then decreases.

(D) The mass $m$ is at height $H$ from point $Q$,where potential energy is taken as zero. From the geometry of the figure,if at some angle $\theta$,the height of mass $m$ above the lowest point $Q$ is $h$,then from $\triangle ABC$,$\cos \theta = \frac{R-h}{R} \Rightarrow h = R(1 - \cos \theta)$. Hence,the potential energy $U(\theta)$ is $U(\theta) = mgh = mgR(1 - \cos \theta)$.
$(b)$ Kinetic energy $K(\theta)$ at position $\theta$ is the loss of potential energy from the initial position $P$ to position $\theta$. Thus,$K(\theta) = mgH - U(\theta) = mgH - mgR(1 - \cos \theta) = mg(H - R(1 - \cos \theta))$.
$(c)$ For $H \ll R$,the motion is simple harmonic with time period $T = 2\pi \sqrt{\frac{R}{g}}$. The time taken to travel from $P$ to $Q$ is one-fourth of this time period. Thus,$t = \frac{T}{4} = \frac{2\pi}{4} \sqrt{\frac{R}{g}} = \frac{\pi}{2} \sqrt{\frac{R}{g}}$.
$(d)$ From energy conservation at the lowest point $Q$,if $m$ has velocity $v$,then $\frac{1}{2}mv^2 = mgH \Rightarrow mv^2 = 2mgH$. The centripetal force is $F_c = \frac{mv^2}{R} = \frac{2mgH}{R}$. The normal reaction $N$ at $Q$ satisfies $N - mg = \frac{mv^2}{R} \Rightarrow N = mg + \frac{2mgH}{R} = mg(1 + \frac{2H}{R})$. This is the force exerted by the block on the surface.

(B) By the principle of conservation of mechanical energy,the potential energy at the initial position equals the kinetic energy at the lowest position.
$m g x(1-\cos \theta) = \frac{1}{2} m v^2$
$v^2 = 2 g x(1-\cos \theta)$
$v = \sqrt{2 g x(1-\cos \theta)}$
For the first position at angle $\theta_1$,$v_1 = \sqrt{2 g x(1-\cos \theta_1)}$.
For the second position at angle $\theta_2$,$v_2 = \sqrt{2 g x(1-\cos \theta_2)}$.
Taking the ratio of the two speeds:
$\frac{v_1}{v_2} = \frac{\sqrt{2 g x(1-\cos \theta_1)}}{\sqrt{2 g x(1-\cos \theta_2)}}$
$\frac{v_1}{v_2} = \sqrt{\frac{1-\cos \theta_1}{1-\cos \theta_2}}$

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$,where $g_{\text{eff}}$ is the effective acceleration due to gravity.
$1$. For a stationary lift,$g_{\text{eff}} = g$,so $T_0 = 2\pi \sqrt{\frac{l}{g}}$.
$2$. When the lift descends at a steady speed,the acceleration is zero. Thus,$g_{\text{eff}} = g$,and the period $T_1 = 2\pi \sqrt{\frac{l}{g}}$. Therefore,$T_0 = T_1$.
$3$. When the lift descends with a constant downward acceleration $a$,a pseudo force $ma$ acts upwards on the pendulum bob. The effective acceleration becomes $g_{\text{eff}} = g - a$. Since $g - a < g$,the new time period $T_2 = 2\pi \sqrt{\frac{l}{g - a}}$ will be greater than $T_0$ and $T_1$.
Thus,$T_0 = T_1 < T_2$.

(B) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.
For a second's pendulum,the time period $T$ is constant $(T = 2 \text{ s})$.
From the formula,we have $T^2 \propto \frac{l}{g}$,which implies $l \propto g$ for a constant $T$.
If the acceleration due to gravity $g'$ on the new planet is $4$ times the acceleration due to gravity $g$ on Earth $(g' = 4g)$,then to keep the time period $T$ constant,the new length $l'$ must satisfy the ratio $\frac{l'}{l} = \frac{g'}{g}$.
Substituting the values,we get $\frac{l'}{l} = \frac{4g}{g} = 4$.
Therefore,the length of the pendulum must be made $4$ times its original length.

(B) 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,it experiences an upward buoyant force (upthrust).
Let $\rho$ be the density of the metal and $\sigma$ be the density of the liquid. Given $\sigma = \frac{\rho}{4}$.
The effective acceleration due to gravity $g_{\text{eff}}$ is given by $g_{\text{eff}} = g(1 - \frac{\sigma}{\rho})$.
Substituting $\sigma = \frac{\rho}{4}$,we get $g_{\text{eff}} = g(1 - \frac{1}{4}) = \frac{3}{4}g$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{\frac{3}{4}g}} = 2\pi \sqrt{\frac{4l}{3g}}$.
This simplifies to $T' = \sqrt{\frac{4}{3}} \times (2\pi \sqrt{\frac{l}{g}}) = \frac{2}{\sqrt{3}}T$.

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
For the first pendulum of length $L_1 = 1.21 \,m$,the time period is $T_1 = 2 \pi \sqrt{\frac{1.21}{g}} = 1.1 \times 2 \pi \sqrt{\frac{1}{g}}$.
For the second pendulum of length $L_2 = 1.0 \,m$,the time period is $T_2 = 2 \pi \sqrt{\frac{1.0}{g}}$.
Thus,$T_1 = 1.1 T_2 = \frac{11}{10} T_2$,which implies $10 T_1 = 11 T_2$.
Let $n_1$ be the number of vibrations of the longer pendulum and $n_2$ be the number of vibrations of the shorter pendulum when they are in phase again.
The condition for being in phase is $n_1 T_1 = n_2 T_2$.
Substituting $T_1 = \frac{11}{10} T_2$,we get $n_1 (\frac{11}{10} T_2) = n_2 T_2$,which simplifies to $11 n_1 = 10 n_2$.
For the smallest integer values,$n_1 = 10$ and $n_2 = 11$.
Therefore,the two pendulums will be in phase again after $10$ vibrations of the longer pendulum.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Initially,$T_1 = 60 \, s$ and length is $l_1$.
When the length is increased by $44 \%$,the new length $l_2 = l_1 + 0.44 l_1 = 1.44 l_1$.
The new time period $T_2$ is given by $T_2 = 2\pi \sqrt{\frac{l_2}{g}}$.
Taking the ratio: $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{1.44} = 1.2$.
Therefore,$T_2 = 1.2 \times T_1 = 1.2 \times 60 \, s = 72 \, s$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the natural logarithm and differentiating,we get $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}$.
Given that the length increases by $0.2 \%$,so $\frac{\Delta L}{L} = 0.002$.
Therefore,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \times 0.002 = 0.001$.
The time lost per day is $\Delta T = \left( \frac{\Delta T}{T} \right) \times t$,where $t = 24 \times 3600 \, s = 86400 \, s$.
Substituting the values,$\Delta T = 0.001 \times 86400 = 86.4 \, s$.

(C) 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,it experiences an upward buoyant force (upthrust).
The effective acceleration due to gravity $g_{\text{eff}}$ is given by $g_{\text{eff}} = g(1 - \frac{\rho_L}{\rho_S})$,where $\rho_L$ is the density of the liquid and $\rho_S$ is the density of the solid (iron).
Given $\rho_L = \frac{1}{12} \rho_S$,we have $\frac{\rho_L}{\rho_S} = \frac{1}{12}$.
Thus,$g_{\text{eff}} = g(1 - \frac{1}{12}) = g(\frac{11}{12})$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{g(\frac{11}{12})}}$.
$T' = T \sqrt{\frac{12}{11}}$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
When a positively charged pendulum is placed in an upward electric field $E$, an electric force $F_e = qE$ acts on the bob in the upward direction.
The effective acceleration due to gravity becomes $g_{eff} = g - \frac{qE}{m}$.
Since $g_{eff} < g$, the denominator in the time period formula decreases.
Therefore, the time period $T$ increases compared to the case without an electric field.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
Initially,the effective acceleration due to gravity is $g = 9.8 \ m/s^2$,so $t = 2\pi \sqrt{\frac{l}{9.8}}$.
When the lift moves upwards with an acceleration $a = 3 \ m/s^2$,the effective acceleration becomes $g' = g + a = 9.8 + 3 = 12.8 \ m/s^2$.
The new time period $t'$ is $t' = 2\pi \sqrt{\frac{l}{12.8}}$.
Dividing the two equations: $\frac{t'}{t} = \frac{2\pi \sqrt{l/12.8}}{2\pi \sqrt{l/9.8}} = \sqrt{\frac{9.8}{12.8}}$.
Therefore,$t' = t \sqrt{\frac{9.8}{12.8}}$.

(D) The maximum tension in the string of a simple pendulum occurs at the lowest point of its oscillation.
At the lowest point,the tension $T$ is given by $T = mg + \frac{mv^2}{l}$,where $v$ is the velocity at the lowest point.
Using the principle of conservation of energy,the potential energy at the maximum angular displacement $\theta_0$ is converted into kinetic energy at the lowest point:
$mgl(1 - \cos \theta_0) = \frac{1}{2}mv^2$
$v^2 = 2gl(1 - \cos \theta_0)$
Substituting $v^2$ into the tension equation:
$T_{\max} = mg + \frac{m(2gl(1 - \cos \theta_0))}{l}$
$T_{\max} = mg + 2mg(1 - \cos \theta_0)$
For small angles,we use the approximation $\cos \theta_0 \approx 1 - \frac{\theta_0^2}{2}$:
$T_{\max} = mg + 2mg(1 - (1 - \frac{\theta_0^2}{2}))$
$T_{\max} = mg + 2mg(\frac{\theta_0^2}{2})$
$T_{\max} = mg(1 + \theta_0^2)$

(D) The time period $T$ 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.
Inside an artificial satellite orbiting the Earth,the satellite and everything inside it are in a state of weightlessness,which means the effective acceleration due to gravity $g_{eff}$ is $0$.
Substituting $g_{eff} = 0$ into the formula,we get $T = 2\pi \sqrt{\frac{l}{0}}$,which implies $T \rightarrow \infty$.
Therefore,the time period of the pendulum is infinity.

(C) The time period $(T)$ of a simple pendulum is given by the formula:
$T = 2\pi \sqrt{\frac{L}{g}}$
Squaring both sides,we get:
$T^2 = \frac{4\pi^2}{g} \times L$
This equation is of the form $y = mx$,where $y = T^2$,$x = L$,and the slope $m = \frac{4\pi^2}{g}$.
Since $m$ is a positive constant,the graph of $T^2$ versus $L$ is a straight line passing through the origin.
Therefore,the correct graph is represented by option $(C)$.

(C) The maximum tension in a simple pendulum occurs at the mean position,given by $T_{max} = mg + \frac{mv^2}{l}$.
Using the conservation of energy,the kinetic energy at the mean position equals the potential energy at the extreme position: $\frac{1}{2}mv^2 = mgl(1 - \cos \theta_0)$,where $\sin \theta_0 = \frac{A}{l} = \frac{10}{100} = 0.1$.
Thus,$\frac{mv^2}{l} = 2mg(1 - \cos \theta_0)$.
Substituting this into the tension formula: $T_{max} = mg + 2mg(1 - \cos \theta_0) = mg(3 - 2\cos \theta_0)$.
Since $\cos \theta_0 = \sqrt{1 - \sin^2 \theta_0} = \sqrt{1 - (0.1)^2} = \sqrt{0.99} \approx 1 - \frac{0.01}{2} = 0.995$.
$T_{max} = 0.25 \times 9.8 \times (3 - 2 \times 0.995) = 2.45 \times (3 - 1.99) = 2.45 \times 1.01 = 2.4745$.
Given $T_{max} = \frac{x}{40}$,we have $x = 40 \times 2.4745 = 98.98 \approx 99$.

(B) At the lowest position,the velocity is $v$. The acceleration is purely centripetal,given by $a_{low} = \frac{v^2}{\ell}$.
At the extreme position,the velocity is zero. The acceleration is purely tangential,given by $a_{ext} = g \sin \theta$.
By conservation of energy between the lowest point and the extreme point:
$\frac{1}{2} mv^2 = mg \ell(1 - \cos \theta) \Rightarrow \frac{v^2}{\ell} = 2g(1 - \cos \theta)$.
Given that the magnitudes of acceleration are equal:
$a_{low} = a_{ext} \Rightarrow \frac{v^2}{\ell} = g \sin \theta$.
Substituting the expression for $\frac{v^2}{\ell}$:
$2g(1 - \cos \theta) = g \sin \theta \Rightarrow 2(1 - \cos \theta) = \sin \theta$.
Using half-angle identities $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$:
$2(2 \sin^2(\theta/2)) = 2 \sin(\theta/2) \cos(\theta/2)$.
Dividing by $2 \sin(\theta/2)$ (assuming $\theta \neq 0$):
$2 \sin(\theta/2) = \cos(\theta/2) \Rightarrow \tan(\theta/2) = \frac{1}{2}$.
Therefore,$\theta = 2 \tan^{-1}\left(\frac{1}{2}\right)$.

(B) The acceleration due to gravity at a height $h$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given $h = 2R$,we have $g' = g \left( \frac{R}{R+2R} \right)^2 = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g'}}$.
For a second's pendulum,$T = 2 \ s$.
Substituting the values: $2 = 2\pi \sqrt{\frac{\ell}{g/9}} = 2\pi \sqrt{\frac{9\ell}{g}}$.
Dividing by $2$: $1 = \pi \sqrt{\frac{9\ell}{g}}$.
Squaring both sides: $1 = \pi^2 \left( \frac{9\ell}{g} \right)$.
Given $g = \pi^2 \ m/s^2$,we substitute this into the equation: $1 = \pi^2 \left( \frac{9\ell}{\pi^2} \right) = 9\ell$.
Therefore,$\ell = \frac{1}{9} \ m$.

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$,where $\ell$ is the length of the pendulum and $g$ is the acceleration due to gravity. The time period is independent of the mass of the bob.
Given that the new length $\ell^{\prime} = \frac{\ell}{2}$.
The new time period $T^{\prime} = 2 \pi \sqrt{\frac{\ell^{\prime}}{g}} = 2 \pi \sqrt{\frac{\ell}{2g}} = \frac{1}{\sqrt{2}} \left( 2 \pi \sqrt{\frac{\ell}{g}} \right) = \frac{1}{\sqrt{2}} T$.
According to the problem,$T^{\prime} = \frac{x}{2} T$.
Therefore,$\frac{x}{2} = \frac{1}{\sqrt{2}}$.
$x = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(D) For a simple pendulum,the angular acceleration $\alpha$ is given by the equation $\alpha = -\omega^2 \theta$,where $\omega^2 = \frac{g}{l}$.
Since the angular accelerations are equal in magnitude,we have $|\alpha_1| = |\alpha_2|$.
Substituting the expression for $\alpha$,we get $\frac{g}{l_1} \theta_1 = \frac{g}{l_2} \theta_2$.
Canceling the acceleration due to gravity $g$ from both sides,we obtain $\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}$.
Rearranging the terms,we get $\theta_1 l_2 = \theta_2 l_1$.

(D) Let the distance between points $A$ and $B$ be $d$. The vertical distance of the bob from the rod is $h = \ell \sin 45^{\circ} = \frac{\ell}{\sqrt{2}}$.
When the bob is displaced by a small angle $\theta$ normal to the plane of the strings,the effective length of the pendulum is the perpendicular distance from the rod to the bob,which is $h = \frac{\ell}{\sqrt{2}}$.
The restoring force for small oscillations is provided by the component of gravity,and the motion is equivalent to a simple pendulum of length $L_{eff} = h = \frac{\ell}{\sqrt{2}}$.
The time period $T$ is given by $T = 2 \pi \sqrt{\frac{L_{eff}}{g}}$.
Substituting $L_{eff} = \frac{\ell}{\sqrt{2}}$,we get $T = 2 \pi \sqrt{\frac{\ell}{\sqrt{2} g}}$.

(A) With respect to the cart,the equilibrium position of the pendulum is shifted due to the pseudo force acting on the bob. The effective acceleration $g_{\text{eff}}$ acting on the pendulum is the vector sum of the acceleration due to gravity $g$ and the pseudo acceleration $a = \sqrt{3}g$ in the opposite direction of the cart's motion.
$g_{\text{eff}} = \sqrt{g^2 + a^2} = \sqrt{g^2 + (\sqrt{3}g)^2} = \sqrt{g^2 + 3g^2} = \sqrt{4g^2} = 2g$.
Given $g = \pi^2 \ m/s^2$,we have $g_{\text{eff}} = 2\pi^2 \ m/s^2$.
The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}}$.
Substituting $L = 50 \ cm = 0.5 \ m$ and $g_{\text{eff}} = 2\pi^2 \ m/s^2$:
$T = 2\pi \sqrt{\frac{0.5}{2\pi^2}} = 2\pi \sqrt{\frac{0.5}{2} \cdot \frac{1}{\pi^2}} = 2\pi \cdot \frac{1}{\pi} \sqrt{0.25} = 2 \cdot 0.5 = 1.0 \ s$.

(A) The time period of a seconds pendulum is $T = 2 \,s$. The formula for the time period is $T = 2 \pi \sqrt{\frac{\ell}{g}}$.
Since $T$ is constant $(2 \,s)$, we have $\ell \propto g$, which implies $\frac{\ell'}{\ell} = \frac{g'}{g}$.
The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
For the planet, $M' = 1.5M$ and $R' = 1.5R$.
Thus, $g' = \frac{G(1.5M)}{(1.5R)^2} = \frac{1.5}{2.25} \frac{GM}{R^2} = \frac{1}{1.5} g$.
Substituting this into the length ratio: $\ell' = \ell \times \frac{g'}{g} = 1 \,m \times \frac{1}{1.5} = \frac{1}{1.5} \,m \approx 0.67 \,m$.

(D) seconds pendulum is a pendulum whose time period is $2 \text{ s}$.
In a space laboratory orbiting the Earth,the laboratory and everything inside it are in a state of weightlessness.
This means the effective acceleration due to gravity $(g_{\text{eff}})$ inside the laboratory is $0$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$.
Substituting $g_{\text{eff}} = 0$ into the formula,we get $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
Therefore,the time period of the pendulum will be infinite.

(D) Let the length of the pendulum be $r$. When the bob is displaced by $90^{\circ}$ from the mean position,its height relative to the lowest point is $r$.
By the law of conservation of energy,the potential energy at the highest point is converted into kinetic energy at the lowest point:
$PE_{top} = KE_{bottom}$
$mgr = \frac{1}{2}mv^2$
$mv^2 = 2mgr$
At the lowest position,the forces acting on the bob are the tension $T$ (upwards) and weight $mg$ (downwards). The net force provides the necessary centripetal force:
$T - mg = \frac{mv^2}{r}$
$T = mg + \frac{mv^2}{r}$
Substituting $mv^2 = 2mgr$ into the equation:
$T = mg + \frac{2mgr}{r} = mg + 2mg = 3mg$.

(C) The electric force acting on the bob is $F_{\text{electric}} = qE$,which acts in the upward direction as per the figure.
The effective weight of the bob is given by $mg_{\text{eff}} = mg - F_{\text{electric}}$.
Therefore,the effective acceleration due to gravity is $g_{\text{eff}} = g - \frac{qE}{m}$.
The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}}$.
Substituting the value of $g_{\text{eff}}$,we get $T = 2 \pi \sqrt{\frac{L}{g - \frac{qE}{m}}} = 2 \pi \left[ \frac{L}{g - \frac{qE}{m}} \right]^{\frac{1}{2}}$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
For a stationary lift,$g_{eff} = g$,so $T_1 = 2\pi \sqrt{\frac{L}{g}} = \sqrt{3} \ s$.
When the lift moves upwards with an acceleration $a = g/3$,the effective acceleration is $g_{eff} = g + a = g + g/3 = 4g/3$.
The new time period $T_2$ is given by $T_2 = 2\pi \sqrt{\frac{L}{4g/3}} = 2\pi \sqrt{\frac{3L}{4g}} = \sqrt{\frac{3}{4}} \times (2\pi \sqrt{\frac{L}{g}})$.
Substituting $T_1 = \sqrt{3} \ s$,we get $T_2 = \frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2} = 1.5 \ s$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
Initially,the point of suspension is at rest,so $g_{eff} = g = 10 \ m/s^2$. Thus,$T_1 = 2\pi \sqrt{\frac{L}{g}}$.
When the point of suspension moves upward with an acceleration $a$,the effective acceleration is $g_{eff} = g + a$.
Given the displacement $y = kt^2$,the acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{d^2y}{dt^2} = \frac{d^2}{dt^2}(kt^2) = 2k$.
Given $k = 1 \ m/s^2$,we have $a = 2(1) = 2 \ m/s^2$.
Therefore,$g_{eff} = g + a = 10 + 2 = 12 \ m/s^2$.
The new time period is $T_2 = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{12}}$.
Now,the ratio $\frac{T_1^2}{T_2^2} = \frac{4\pi^2 (L/g)}{4\pi^2 (L/g_{eff})} = \frac{g_{eff}}{g} = \frac{12}{10} = \frac{6}{5}$.

(B) 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,we can see that $T \propto \sqrt{L}$.
Let the original length be $L_1 = L$ and the original time period be $T_1 = T$.
The new length is $L_2 = 3L$.
Let the new time period be $T_2$.
Using the proportionality $T \propto \sqrt{L}$,we have: $\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}$.
Substituting the values: $\frac{T_2}{T} = \sqrt{\frac{3L}{L}} = \sqrt{3}$.
Therefore,$T_2 = \sqrt{3} T$.

(D) small sphere oscillating in a watch glass acts as a simple pendulum.
The effective length $L$ of this equivalent pendulum is equal to the radius of curvature $R$ of the watch glass.
Given,$R = 1.6 \ m$ and $g = 10 \ m/s^2$.
The formula for the time period $T$ of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.
Substituting the values,we get $T = 2\pi \sqrt{\frac{1.6}{10}}$.
$T = 2\pi \sqrt{0.16}$.
$T = 2\pi \times 0.4$.
$T = 0.8\pi \ s$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
When the lift is at rest,$g_{eff} = g$,so $T = 2\pi \sqrt{\frac{L}{g}}$.
When the lift accelerates upward with acceleration $a$,the effective gravity becomes $g_{eff} = g + a$.
The new time period is $T' = 2\pi \sqrt{\frac{L}{g+a}}$.
Given that $T' = \frac{T}{2}$,we have $\frac{T}{2} = 2\pi \sqrt{\frac{L}{g+a}}$.
Substituting $T = 2\pi \sqrt{\frac{L}{g}}$,we get $\frac{1}{2} \left( 2\pi \sqrt{\frac{L}{g}} \right) = 2\pi \sqrt{\frac{L}{g+a}}$.
Squaring both sides: $\frac{1}{4} \left( \frac{L}{g} \right) = \frac{L}{g+a}$.
This simplifies to $g + a = 4g$,which gives $a = 3g$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Since $T \propto \sqrt{l}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$.
Given that the periodic time increases by $20 \%$,the new time period $T_2$ is $T_2 = T_1 + 0.20 T_1 = 1.2 T_1$.
Therefore,$\frac{T_2}{T_1} = 1.2$.
Squaring both sides,we get $\left(\frac{T_2}{T_1}\right)^2 = \frac{l_2}{l_1}$.
Substituting the values,$\frac{l_2}{l_1} = (1.2)^2 = 1.44$.

(A) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Given for the planet: $M_p = 2M_e$ and $R_p = 2R_e$.
Thus,the ratio of gravity on Earth to the planet is:
$\frac{g_e}{g_p} = \frac{M_e}{M_p} \times \left(\frac{R_p}{R_e}\right)^2 = \frac{M_e}{2M_e} \times \left(\frac{2R_e}{R_e}\right)^2 = \frac{1}{2} \times 4 = 2$.
So,$g_p = \frac{g_e}{2}$.
The time period of a pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.
Therefore,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{2}$.
Since the time period of a second's pendulum on Earth is $T_e = 2 \ s$,we have:
$T_p = T_e \times \sqrt{2} = 2\sqrt{2} \ s$.

(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
When the lift accelerates downwards with an acceleration $a = \frac{g}{4}$,the effective acceleration due to gravity $g_{\text{eff}}$ becomes $g - a$.
$g_{\text{eff}} = g - \frac{g}{4} = \frac{3g}{4}$.
The new time period $T_1$ is given by $T_1 = 2 \pi \sqrt{\frac{L}{g_{\text{eff}}}} = 2 \pi \sqrt{\frac{L}{\frac{3g}{4}}}$.
$T_1 = 2 \pi \sqrt{\frac{4L}{3g}} = 2 \pi \sqrt{\frac{L}{g}} \times \sqrt{\frac{4}{3}}$.
Since $T = 2 \pi \sqrt{\frac{L}{g}}$,we have $T_1 = T \times \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} T$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
This implies $T \propto \frac{1}{\sqrt{g}}$.
When the bob is immersed in a liquid,the effective acceleration due to gravity $g'$ changes due to the buoyant force.
The effective weight $W' = V \rho g - V \sigma g$,where $\rho$ is the density of the bob and $\sigma$ is the density of the liquid.
Given $\sigma = \frac{\rho}{8}$,the effective acceleration $g'$ is:
$g' = g \left(1 - \frac{\sigma}{\rho}\right) = g \left(1 - \frac{1}{8}\right) = g \left(\frac{7}{8}\right)$.
The new time period $T'$ is given by:
$T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g(7/8)}} = \sqrt{\frac{8}{7}} \left(2\pi \sqrt{\frac{l}{g}}\right) = \sqrt{\frac{8}{7}} T$.

(A) For a simple pendulum with small angular displacement $\theta$,the angular position $\theta(t)$ at time $t$ is given by the equation of Simple Harmonic Motion $(SHM)$: $\theta(t) = \theta_0 \cos(\omega t)$.
Here,the initial angular displacement is $\theta$,so $\theta(t) = \theta \cos(\omega t)$.
The angular frequency $\omega$ of a simple pendulum is given by $\omega = \sqrt{\frac{g}{L}}$.
Therefore,the angular displacement at time $t$ is $\theta(t) = \theta \cos\left(\sqrt{\frac{g}{L}} \cdot t\right)$.
The linear displacement $s$ of the bob along the arc is given by the product of the length of the pendulum and the angular displacement: $s = L \cdot \theta(t)$.
Substituting the expression for $\theta(t)$,we get $s = L \theta \cos\left(\sqrt{\frac{g}{L}} \cdot t\right)$.

(D) The frequency of a simple pendulum is given by the formula: $n = \frac{1}{2 \pi} \sqrt{\frac{g}{L}}$.
From this relation,we can see that the frequency $n$ is inversely proportional to the square root of the length $L$,i.e.,$n \propto \frac{1}{\sqrt{L}}$.
Therefore,the ratio of the frequencies is given by: $\frac{n_1}{n_2} = \sqrt{\frac{L_2}{L_1}}$.
Given the ratio of frequencies $\frac{n_1}{n_2} = \frac{3}{2}$,we square both sides to find the ratio of lengths:
$(\frac{n_1}{n_2})^2 = \frac{L_2}{L_1} \Rightarrow (\frac{3}{2})^2 = \frac{L_2}{L_1} \Rightarrow \frac{9}{4} = \frac{L_2}{L_1}$.
Thus,the ratio of the lengths $L_1 : L_2$ is $4 : 9$.

(C) The frequency of a simple pendulum is given by $f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$,which implies $f \propto \frac{1}{\sqrt{l}}$.
Let the original length be $l_1 = l$ and the original frequency be $f_1 = f$.
The length is increased by three times its original length,so the new length $l_2 = l + 3l = 4l$.
Using the relation $\frac{f_2}{f_1} = \sqrt{\frac{l_1}{l_2}}$,we get $\frac{f_2}{f} = \sqrt{\frac{l}{4l}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the new frequency is $f_2 = \frac{f}{2}$.