SHM of Simple Pendulum Questions in English

(D) At the mean position,the tension $T$ is given by $T = mg + \frac{mv^2}{l}$,where $l$ is the length of the string.
Given that $T = 3mg$,we have $mg + \frac{mv^2}{l} = 3mg$,which simplifies to $\frac{mv^2}{l} = 2mg$,so $v^2 = 2gl$.
Let $\theta$ be the maximum angular displacement from the vertical. By the principle of conservation of energy,the kinetic energy at the mean position must equal the potential energy at the maximum displacement: $\frac{1}{2}mv^2 = mgl(1 - \cos \theta)$.
Substituting $v^2 = 2gl$,we get $\frac{1}{2}m(2gl) = mgl(1 - \cos \theta)$.
This simplifies to $mgl = mgl(1 - \cos \theta)$,which implies $1 = 1 - \cos \theta$,so $\cos \theta = 0$.
Therefore,$\theta = 90^\circ$.

(C) The acceleration of the bob of a simple pendulum in motion consists of two components:
$1$. Centripetal acceleration $(a_c)$: Directed towards the point of suspension along the string.
$2$. Tangential acceleration $(a_t)$: Directed along the tangent to the circular path of the bob.
The net acceleration $\vec a$ is the vector sum of these two components: $\vec a = \vec a_c + \vec a_t$.
Since the bob is moving in a circular arc,the resultant acceleration vector $\vec a$ must point towards the interior of the arc,specifically between the string and the tangent,as shown in option $(C)$.

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

(D) 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.
For the period $T$ to decrease,the effective acceleration $g_{eff}$ must increase.
When a rocket moves up with a uniform acceleration $a$,the effective acceleration experienced by the pendulum is $g_{eff} = g + a$.
Since $g + a > g$,the effective gravity increases,which leads to a decrease in the time period $T$.
Therefore,the correct option is $D$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the lift moves upward with acceleration $a = \frac{g}{4}$,the effective acceleration due to gravity becomes $g' = g + a = g + \frac{g}{4} = \frac{5g}{4}$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{5g/4}} = 2\pi \sqrt{\frac{4l}{5g}}$.
Comparing $T'$ with $T$,we get $\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{5g/4}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
Therefore,$T' = \frac{2T}{\sqrt{5}}$.

(C) 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 lift is at rest,the effective acceleration is $g_{eff} = g$. Therefore,$T = 2\pi \sqrt{\frac{l}{g}}$.
When the resultant acceleration becomes $g' = g/4$,the new time period $T'$ is given by:
$T' = 2\pi \sqrt{\frac{l}{g/4}}$
$T' = 2\pi \sqrt{\frac{4l}{g}}$
$T' = 2 \times (2\pi \sqrt{\frac{l}{g}})$
$T' = 2T$.

(B) According to the law of conservation of energy,the kinetic energy at the mean position is equal to the potential energy at the extreme position.
$1/2 mv^2 = mgh$
$v = \sqrt{2gh}$
Given,$h = 10 \, cm = 0.1 \, m$ and $g = 9.8 \, m/s^2$.
$v = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4 \, m/s$.

(D) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
In 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 $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
Therefore,the time period of a simple pendulum on a freely moving artificial satellite is infinite.

(B) The time period of a pendulum clock is given by $T = 2\pi \sqrt{l/g}$. In an orbiting satellite,the astronaut experiences weightlessness,which means the effective acceleration due to gravity $g$ is $0$. As $g \to 0$,$T \to \infty$,so a pendulum clock will not work.
However,a watch that uses a main spring to keep it going relies on the elastic properties of the spring and the oscillation of a balance wheel,which are independent of the gravitational field. Therefore,it will function correctly in a satellite.

(B) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} = \frac{1}{2} \alpha \Delta \theta$.
Given $\Delta \theta = t$,the fractional change is $\frac{\Delta T}{T} = \frac{1}{2} \alpha t$.
The total number of seconds in a day is $86400 \text{ s}$.
Therefore,the loss in time per day is $\Delta T_{day} = \left( \frac{\Delta T}{T} \right) \times 86400 = \frac{1}{2} \alpha t \times 86400$.

(A) The displacement of the oscillator is given by $x = A \cos \left( \omega t + \frac{\pi}{4} \right)$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -A\omega \sin \left( \omega t + \frac{\pi}{4} \right)$.
For the speed to be maximum,the magnitude of the sine function must be $1$,i.e.,$|\sin \left( \omega t + \frac{\pi}{4} \right)| = 1$.
This occurs when $\omega t + \frac{\pi}{4} = \frac{\pi}{2}$.
Solving for $t$: $\omega t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
Therefore,$t = \frac{\pi}{4\omega}$.

(D) For a simple harmonic oscillator,the kinetic energy $T$ at displacement $X$ is given by $T = \frac{1}{2} m \omega^2 (a^2 - X^2)$.
The potential energy $V$ at displacement $X$ is given by $V = \frac{1}{2} m \omega^2 X^2$.
To find the ratio of kinetic energy $T$ to potential energy $V$,we divide the two expressions:
$\frac{T}{V} = \frac{\frac{1}{2} m \omega^2 (a^2 - X^2)}{\frac{1}{2} m \omega^2 X^2}$.
Canceling the common terms $\frac{1}{2} m \omega^2$,we get:
$\frac{T}{V} = \frac{a^2 - X^2}{X^2}$.

(B) The motion of a ball dropped in a tunnel through the center of the Earth is simple harmonic motion $(S.H.M.)$.
The time period of this oscillation is given by $T = 2\pi \sqrt{\frac{R}{g}}$, where $R$ is the radius of the Earth and $g$ is the acceleration due to gravity.
Substituting the values $R \approx 6400 \; km$ and $g \approx 9.8 \; m/s^2$, we get $T \approx 84.6 \; \text{minutes}$.
The time taken to travel from one end to the other end of the tunnel is half of the total time period.
Therefore, $t = \frac{T}{2} = \frac{84.6}{2} = 42.3 \; \text{minutes}$.

(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 relation,we can see that $T \propto \sqrt{l}$.
If the period $T$ is to be doubled $(2T)$,then the new length $l'$ must satisfy $2T \propto \sqrt{l'}$.
Dividing the two relations: $\frac{2T}{T} = \frac{\sqrt{l'}}{\sqrt{l}} \implies 2 = \sqrt{\frac{l'}{l}}$.
Squaring both sides,we get $4 = \frac{l'}{l}$,which means $l' = 4l$.
Therefore,the length must be made four times the original length.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the effective length of the pendulum.
The effective length $L$ is the distance from the point of suspension to the center of gravity (c.g.) of the bob.
Initially, the center of gravity of the mercury-filled sphere is at its geometric center.
When a little mercury is drained off, the level of mercury inside the sphere drops, causing the center of gravity of the bob to shift downwards (away from the point of suspension).
As the center of gravity moves downwards, the effective length $L$ of the pendulum increases.
Since $T \propto \sqrt{L}$, an increase in $L$ leads to an increase in the time period $T$ of the pendulum.

(B) Initially,the time period of the simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the train accelerates with a uniform acceleration $a$,a pseudo force acts on the pendulum bob in the opposite direction of the acceleration.
The effective acceleration due to gravity $(g_{\text{eff}})$ becomes the vector sum of the gravitational acceleration $g$ and the pseudo acceleration $a$. Thus,$g_{\text{eff}} = \sqrt{g^2 + a^2}$.
Since $g_{\text{eff}} > g$,the new time period $T' = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$ will be less than the initial time period $T$.
Therefore,the period of oscillation decreases.

(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, we can observe the following:
$1$. $T$ is directly proportional to $\sqrt{L}$.
$2$. $T$ is inversely proportional to $\sqrt{g}$.
$3$. $T$ is independent of the mass, size, and material of the bob.
$4$. For small amplitudes, $T$ is independent of the amplitude.
Therefore, the statement that the period is dependent on the mass, size, and material of the bob is false.

(C) In a stationary lift,the time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the lift accelerates upwards with an acceleration $a = \frac{g}{4}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{4} = \frac{5g}{4}$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g_{eff}}} = 2\pi \sqrt{\frac{l}{5g/4}} = 2\pi \sqrt{\frac{4l}{5g}}$.
We can rewrite this as $T' = \sqrt{\frac{4}{5}} \times (2\pi \sqrt{\frac{l}{g}}) = \frac{2}{\sqrt{5}} T$.

(D) When a trolley moves with a horizontal acceleration $a$,a pseudo force $ma$ acts on the bob of the pendulum in the direction opposite to the acceleration of the trolley.
The effective acceleration due to gravity $g'$ is the vector sum of the acceleration due to gravity $g$ (acting downwards) and the pseudo acceleration $a$ (acting horizontally).
Since these two vectors are perpendicular to each other,the magnitude of the effective acceleration $g'$ is given by:
$g' = \sqrt{g^2 + a^2}$
Thus,the correct option is $D$.

(D) second's pendulum is a pendulum whose time period is $2 \, s$ on the surface of the Earth.
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 $(g_{eff})$ inside the orbiting laboratory is $0$.
The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
Substituting $g_{eff} = 0$ into the formula,we get $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
Therefore,the time period of the pendulum is infinite.

(B) The total energy $E$ of a simple harmonic oscillator is given by $E = \frac{1}{2}mv_{\max}^2$.
From this,we can express the maximum velocity $v_{\max}$ as $v_{\max} = \sqrt{\frac{2E}{m}}$.
The maximum linear momentum $p_{\max}$ is defined as $p_{\max} = m \cdot v_{\max}$.
Substituting the expression for $v_{\max}$,we get $p_{\max} = m \cdot \sqrt{\frac{2E}{m}} = \sqrt{m^2 \cdot \frac{2E}{m}} = \sqrt{2mE}$.
Therefore,the maximum linear momentum is $\sqrt{2mE}$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
For a seconds pendulum,the time period $T$ is fixed at $2\, s$ on both the Earth and the Moon.
Since $T = 2\pi \sqrt{\frac{l}{g}}$,we have $T^2 = 4\pi^2 \frac{l}{g}$,which implies $l = \frac{T^2 g}{4\pi^2}$.
Since $T$ is constant,$l \propto g$.
Therefore,$\frac{l_m}{l_e} = \frac{g_m}{g_e}$.
Given $g_m = \frac{1}{6} g_e$ and $l_e = 1\, m$,we get $l_m = \frac{1}{6} \times 1\, m = \frac{1}{6}\, m$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,which implies $T \propto \sqrt{l}$.
Taking the derivative,we get $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.
Given that the length is decreased by $2\%$,we have $\frac{\Delta l}{l} = 0.02$.
Therefore,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \times 0.02 = 0.01$.
This means the pendulum loses $0.01$ seconds for every second of its time period.
Since a second's pendulum has a time period of $T = 2 \text{ seconds}$,the loss per oscillation is $\Delta T = 0.01 \times 2 = 0.02 \text{ seconds}$.
The number of oscillations in one day $(86400 \text{ seconds})$ is $N = \frac{86400}{2} = 43200$.
The total time lost per day is $N \times \Delta T = 43200 \times 0.02 = 864 \text{ seconds}$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
In a stationary lift,$g_{eff} = g$,so $T = 2\pi \sqrt{\frac{L}{g}}$.
When the lift moves upwards with an acceleration $a = 5g$,the effective acceleration becomes $g_{eff} = g + a = g + 5g = 6g$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{L}{6g}}$.
Comparing the two,we get $T' = \frac{T}{\sqrt{6}}$.
Since $\frac{T}{\sqrt{6}}$ is not listed in the options,the correct choice is $(d)$.

(B) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.
From this relation,we can see that $T \propto \sqrt{l}$.
Taking the logarithmic derivative,we get $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.
Given that the length is increased by $1\%$,we have $\frac{\Delta l}{l} = 1\%$.
Substituting this value,we get $\frac{\Delta T}{T} = \frac{1}{2} \times 1\% = 0.5\%$.
Therefore,the time period will increase by $0.5\%$.

(B) At point $A$ (or $C$),the bob is at its maximum height $H$ relative to point $B$,so its kinetic energy is zero and potential energy is $mgH$.
At point $B$,the bob is at its lowest position,so its potential energy is zero and all energy is converted into kinetic energy.
According to the law of conservation of mechanical energy:
$PE_{A} + KE_{A} = PE_{B} + KE_{B}$
$mgH + 0 = 0 + \frac{1}{2}mv^2$
$mgH = \frac{1}{2}mv^2$
$v^2 = 2gH$
$v = \sqrt{2gH}$

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
For a given angular amplitude $\theta_0$,the maximum velocity $v_{max}$ is given by $v_{max} = \omega A = \sqrt{\frac{g}{L}} \cdot (L \theta_0) = \theta_0 \sqrt{gL}$.
As the length $L$ increases,the maximum velocity $v_{max}$ increases,not decreases. Thus,$(c)$ is incorrect.
Option $(a)$ is incorrect because the frequency $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$ is independent of mass.
Option $(b)$ is incorrect because the tension at any angle $\theta$ is $T = Mg \cos \theta + \frac{Mv^2}{L}$. Since $\frac{Mv^2}{L} > 0$,the tension is always greater than $Mg \cos \theta$.
Option $(d)$ is correct. The time period is $T = 2\pi \sqrt{\frac{L}{g}}$. Taking logs and differentiating,$\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} = \frac{1}{2} \alpha \Delta \theta$. This is independent of the length $L$.

(C) When the bob is displaced by an angle $\theta$,it rises to a vertical height $h$ above its mean position.
By the law of conservation of mechanical energy,the potential energy at the extreme position is converted into kinetic energy at the mean (lowest) position.
$mgh = \frac{1}{2}mv_{\max}^2$
$v_{\max} = \sqrt{2gh}$
From the geometry of the pendulum,the vertical distance from the point of suspension to the bob at the extreme position is $l \cos \theta$.
Therefore,the height $h$ is given by:
$h = l - l \cos \theta = l(1 - \cos \theta)$
Substituting the value of $h$ into the velocity equation:
$v_{\max} = \sqrt{2gl(1 - \cos \theta)}$

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
When a pendulum falls freely,the acceleration due to gravity $g$ is exactly cancelled by the pseudo-force acting on the bob due to the downward acceleration of the support.
Therefore,the effective acceleration due to gravity $g_{eff} = g - g = 0$.
Substituting this into the formula,we get $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
Since the time period becomes infinite,the frequency $f = \frac{1}{T} = 0$.
This implies that the pendulum does not oscillate at all.

(D) Let the velocity of the bob be $v$ at point $B$,where the string makes an angle of $60^o$ with the vertical. Using the law of conservation of mechanical energy between the lowest point $A$ and point $B$:
$KE_A + PE_A = KE_B + PE_B$
Taking the lowest point $A$ as the reference level for potential energy $(PE_A = 0)$:
$\frac{1}{2}mv_A^2 + 0 = \frac{1}{2}mv^2 + mgh$
Here,$h = l(1 - \cos \theta)$ is the height of point $B$ above point $A$.
Given $v_A = 3\, m/s$,$l = 0.5\, m$,$\theta = 60^o$,and $g = 10\, m/s^2$:
$\frac{1}{2} \times m \times 3^2 = \frac{1}{2} \times m \times v^2 + m \times 10 \times 0.5 \times (1 - \cos 60^o)$
$4.5 = 0.5v^2 + 5 \times (1 - 0.5)$
$4.5 = 0.5v^2 + 5 \times 0.5$
$4.5 = 0.5v^2 + 2.5$
$0.5v^2 = 2$
$v^2 = 4$
$v = 2\, m/s$

(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 formula,we can see that $T \propto \sqrt{l}$.
Let the initial length be $l_1 = l$ and the initial time period be $T_1 = 2 \, sec$.
Let the new length be $l_2 = 4l$ and the new time period be $T_2$.
Using the ratio method: $\frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}}$.
Substituting the values: $\frac{2}{T_2} = \sqrt{\frac{l}{4l}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$T_2 = 2 \times 2 = 4 \, sec$.

(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.
As seen from the formula, the time period $T$ depends only on the length of the pendulum and the acceleration due to gravity.
It is independent of the mass, material, or density of the bob.
Therefore, replacing the metal bob with a wooden bob will not change the time period of the pendulum.

(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:
$T^2 = 4\pi^2 \frac{l}{g}$
Rearranging the terms to isolate the ratio of length to the square of the time period:
$\frac{l}{T^2} = \frac{g}{4\pi^2}$
Since $g$ (acceleration due to gravity) and $\pi$ are constants,the ratio $\frac{l}{T^2}$ is a constant.
Therefore,the correct option is $C$.

(D) 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 is inversely proportional to the square root of the acceleration due to gravity: $T \propto \frac{1}{\sqrt{g}}$.
Let $T_e$ and $g_e$ be the time period and acceleration due to gravity on Earth,and $T_p$ and $g_p$ be the time period and acceleration due to gravity on the other planet.
Given that the acceleration due to gravity on the planet is half that of the Earth,we have $g_p = \frac{g_e}{2}$.
Using the proportionality relation:
$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{\frac{g_e}{g_e/2}} = \sqrt{2}$.
Therefore,the new time period $T_p = \sqrt{2}T_e = \sqrt{2}T$.

(A) 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 $l_1 = 100$ units. Then the new length $l_2 = 100 + 21 = 121$ units.
The ratio of the time periods is $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{\frac{121}{100}} = \frac{11}{10} = 1.1$.
Thus,$T_2 = 1.1 T_1$.
The percentage increase in the time period is given by $\frac{T_2 - T_1}{T_1} \times 100 = \frac{1.1 T_1 - T_1}{T_1} \times 100 = 0.1 \times 100 = 10\%$.
Therefore,the correct option is $A$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,which implies $T \propto \sqrt{l}$.
Let the initial length be $l_1 = l$ and the initial time period be $T_1$.
If the length is increased by $300\%$,the new length $l_2 = l + 300\% \text{ of } l = l + 3l = 4l$.
Using the proportionality $T \propto \sqrt{l}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{\frac{4l}{l}} = \sqrt{4} = 2$.
Thus,$T_2 = 2T_1$.
The percentage increase in the time period is given by $\frac{T_2 - T_1}{T_1} \times 100\% = \frac{2T_1 - T_1}{T_1} \times 100\% = 100\%$.
Therefore,the time period increases by $100\%$.

(B) seconds pendulum is a pendulum whose period of oscillation is exactly $2 \ s$.
Using the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides: $T^2 = 4\pi^2 \frac{l}{g}$.
Rearranging for length $l$: $l = \frac{gT^2}{4\pi^2}$.
Given $T = 2 \ s$ and taking $g = 9.8 \ m/s^2$ (or $980 \ cm/s^2$) and $\pi^2 \approx 9.87$:
$l = \frac{980 \times (2)^2}{4 \times 9.87} = \frac{980 \times 4}{39.48} \approx 99.29 \ cm$.
Standard approximation often used in physics textbooks takes $g = \pi^2 \ m/s^2$,which yields $l = 1 \ m = 100 \ cm$. However,using $g = 9.8 \ m/s^2$,the value is approximately $99.3 \ cm$. Among the given options,$99 \ cm$ is the closest standard value.

(D) When a lift descends with an acceleration $a$,the effective acceleration due to gravity $g_{eff}$ experienced by the pendulum is $g_{eff} = g - a$.
In this case,the lift is descending with an acceleration $a = g$.
Therefore,the effective acceleration $g_{eff} = g - g = 0$.
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
Substituting $g_{eff} = 0$,we get $T = 2\pi \sqrt{\frac{l}{0}} = \infty$.
Thus,the time period becomes infinite.

(D) 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 center of mass of the oscillating body).
When the chimpanzee stands up,the center of mass of the system shifts upward,closer to the point of suspension.
This results in a decrease in the effective length $l$ of the swing.
Since $T \propto \sqrt{l}$,a decrease in $l$ leads to a decrease in the time period $T$.

(C) The formula for the time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Given: length $l = 1 \, m$ and acceleration due to gravity $g = \pi^2 \, m/s^2$.
Substituting these values into the formula:
$T = 2\pi \sqrt{\frac{1}{\pi^2}}$
$T = 2\pi \times \frac{1}{\pi}$
$T = 2 \, s$.
Therefore,the time period is $2 \, s$.

(C) The time period 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.
As observed from the formula,the time period $T$ depends only on the length of the pendulum and the acceleration due to gravity.
It is independent of the mass of the bob or the object attached to the pendulum.
Therefore,placing another plate on the first plate does not change the mass-independent nature of the oscillation period.
Thus,the time period will remain the same.

(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 and $g$ is the acceleration due to gravity.
From this formula,it is clear that $T \propto \sqrt{l}$.
The time period 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 values: $\frac{2T}{T} = \sqrt{\frac{l_2}{l}}$,which simplifies to $2 = \sqrt{\frac{l_2}{l}}$.
Squaring both sides,we get $4 = \frac{l_2}{l}$,which implies $l_2 = 4l$.

(B) 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.
As seen from the formula,the time period $T$ depends only on the length of the pendulum and the acceleration due to gravity.
It is independent of the mass of the bob.
$A$ seconds pendulum is defined as a pendulum with a time period of $2\, s$.
Since the time period does not change with the mass of the bob,the new time period remains $2\, s$.

(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.
On the surface of the moon,the acceleration due to gravity $g_m$ is approximately $1/6$th of the acceleration due to gravity on Earth $(g_e)$.
Since $T \propto \frac{1}{\sqrt{g}}$,a decrease in the value of $g$ leads to an increase in the time period $T$.
Therefore,the time period of a simple pendulum increases when it oscillates on the surface of the moon.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
When the lift is stationary,$g_{eff} = g$,so $T = 2\pi \sqrt{\frac{L}{g}}$.
When the lift falls freely,it experiences weightlessness,meaning the effective acceleration due to gravity $g_{eff} = 0$.
Substituting $g_{eff} = 0$ into the formula,the time period $T' = 2\pi \sqrt{\frac{L}{0}} = \infty$.
Since frequency $f = \frac{1}{T'}$,we get $f = \frac{1}{\infty} = 0$.
Therefore,the frequency of oscillation is zero.

(D) 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 van moves with a uniform velocity,its acceleration is zero.
Therefore,the net force acting on the pendulum bob remains the same as when the van was stationary.
Since the effective acceleration due to gravity $g_{eff}$ remains equal to the acceleration due to gravity $g$,the time period $T$ remains unchanged.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Given that the length is increased by $44\%$,the new length $l_2 = l_1 + 0.44 l_1 = 1.44 l_1$.
Since $T \propto \sqrt{l}$,the ratio of the new time period $T_2$ to the original time period $T_1$ is:
$\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{\frac{1.44 l_1}{l_1}} = \sqrt{1.44} = 1.2$.
This implies $T_2 = 1.2 T_1$.
The percentage change in the time period is given by:
$\text{Percentage change} = \left( \frac{T_2 - T_1}{T_1} \right) \times 100 = \left( \frac{1.2 T_1 - T_1}{T_1} \right) \times 100 = 0.2 \times 100 = 20\%$.
Therefore,the time period increases by $20\%$.