Question

A simple pendulum has a time period of $3.0{\rm{ s}}$. If the point of suspension of the pendulum starts moving vertically upward with a velocity $\upsilon = Kt$ where $K = 4.4{\rm{ m}}{{\rm{s}}^{ - 2}}$, the new time period will be (Take $g = 10{\rm{ m}}{{\rm{s}}^{ - 2}}$)
(1) $\dfrac{9}{4}s$
(2) $\dfrac{5}{3}s$
(3) $2.5s$
(4) $4.4s$

Answer 1

Hint: For a simple pendulum in an oscillating motion having the length of the string of the pendulum be $l$ and the mass of the bob be $m$, then the time period $T$ of the pendulum is given by this relation-
$T = 2\pi \sqrt {\dfrac{l}{g}}$
Where, $g$ is the acceleration due to gravity.

Complete step by step answer:
Given:
The time period of given simple pendulum at suspended position ${T_1} = 3.0{\rm{ s}}$
Also, from the formula for the time period of the pendulum we have,
${T_1} = 2\pi \sqrt {\dfrac{l}{g}}$
Substituting, ${T_1} = 3.0{\rm{ s}}$ and $g = 10{\rm{ m}}{{\rm{s}}^{ - 2}}$ in the formula we get,
$\begin{array}{l} 3.0 = 2\pi \sqrt {\dfrac{l}{{10}}} \\\ \dfrac{{3.0}}{{2\pi }} = \sqrt {\dfrac{l}{{10}}} \end{array}$
Squaring both sides and solving we get,
$\begin{array}{l} {\left( {\dfrac{{3.0}}{{2\pi }}} \right)^2} = \dfrac{l}{{10}}\\\ l = {\left( {\dfrac{{3.0}}{{2\pi }}} \right)^2} \times 10\\\ l = 2.28{\rm{ m}} \end{array}$
So, the length of the string of the pendulum is $2.28{\rm{ m}}$.
The expression for the upward velocity of the pendulum is given as-
$\upsilon = Kt$
Where, $K = 4.4{\rm{ m}}{{\rm{s}}^{ - 2}}$ and $t$ is the instantaneous time period.
Now, since the pendulum is moving upwards with a velocity $\upsilon$ the acceleration in this direction will be-
$a = \dfrac{{d\upsilon }}{{dt}}$
Substituting the value of $\upsilon$ in the expression we get,
$a = \dfrac{d}{{dt}}\left( {Kt} \right)$
We can write this as-
$\begin{array}{l} a = K\dfrac{{dt}}{{dt}}\\\ {\rm{or}}\\\ a = K \end{array}$
We know the value of $K$, so substituting $K = 4.4{\rm{ m}}{{\rm{s}}^{ - 2}}$ we get,
$a = 4.4{\rm{ m}}{{\rm{s}}^{ - 2}}$
So, for this position of the pendulum the total effective acceleration due to gravity becomes, ${g_{eff}} = \left( {g + a} \right)$
We know that, $g = 10{\rm{ m}}{{\rm{s}}^{ - 2}}$ and $a = 4.4{\rm{ m}}{{\rm{s}}^{ - 2}}$, substituting the value we get,
$\begin{array}{l} {g_{eff}} = \left( {10 + 4.4} \right)\\\ {g_{eff}} = 14.4{\rm{ m}}{{\rm{s}}^{ - 2}} \end{array}$
Now using the formula for the new period of the pendulum we get,
${T_2} = 2\pi \sqrt {\dfrac{l}{{{g_{eff}}}}}$
Substituting the values $l = 2.28{\rm{ m}}$ and ${g_{eff}} = 14.4{\rm{ m}}{{\rm{s}}^{ - 2}}$ in the formula we get,
$\begin{array}{l} {T_2} = 2\pi \sqrt {\dfrac{{2.28}}{{14.4}}} \\\ \Rightarrow {T_2} = 2\pi \times 0.398\\\ \Rightarrow {T_2} = 2.5{\rm{ s}} \end{array}$
Therefore, the new period of the pendulum is $2.5{\rm{ s}}$ and the correct answer is (3) $2.5s$

Note: It should be noted that the value of acceleration $a$ used in calculating the effective acceleration due to gravity ${g_{eff}}$ has same value of magnitude but the direction is opposite to the direction of motion of the pendulum i.e. the direction of $a$ is downwards.