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Physics-

General

Easy

Question

The angular amplitude of a simple pendulum is $theta subscript 0 end subscript$ . The maximum tension in its string will be

$m g (1 - θ_{0})$
$m g (1 + θ_{0})$
$m g (1 - θ_{0}^{2})$
$m g (1 + θ_{0}^{2})$

The correct answer is: $m g open parentheses 1 plus theta subscript 0 end subscript superscript 2 end superscript close parentheses$

The simple pendulum at angular amplitude $theta subscript 0 end subscript$ is shown in the figure.
Maximum tension in the string is
$T subscript m a x end subscript equals m g plus fraction numerator m v to the power of 2 end exponent over denominator l end fraction$ (i)
When bob of the pendulum comes from $A$ to $B$ , it covers a vertical distance $h$

$therefore cos invisible function application theta subscript 0 end subscript equals fraction numerator l minus h over denominator l end fraction$
$rightwards double arrow h equals l left parenthesis 1 minus cos invisible function application theta subscript 0 end subscript right parenthesis$ (ii)
Also during $A$ to $B$ , potential energy of bob converts into kinetic energy $i e$ , $m g h equals fraction numerator 1 over denominator 2 end fraction blank m v to the power of 2 end exponent$
$therefore v equals square root of 2 g h end root$ (iii)
Thus, using Eqs. (i),(ii) and (iii), we obtain
$T subscript m a x end subscript equals m g plus fraction numerator 2 m g over denominator l end fraction l left parenthesis 1 minus cos invisible function application theta right parenthesis$
$equals m g plus 2 m g open square brackets 1 minus 1 plus fraction numerator theta subscript 0 end subscript superscript 2 end superscript over denominator 2 end fraction close square brackets$
$equals m g left parenthesis 1 plus theta subscript 0 end subscript superscript 2 end superscript right parenthesis$

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