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Question : 10 of 26

Marks: +1, -0

The bob of a simple pendulum of length 200 cm is released from horizontal position. If

10 \%

of its initial energy is lost due to air resistance, then the speed of bob at the mean position is

(Acceleration due to gravity

=10\,\mathrm{m}\,\mathrm{s}^{-2}

[AP EAPCET 21 May 2025 S2]

$6\,\mathrm{m}\,\mathrm{s}^{-1}$
$3\,\mathrm{m}\,\mathrm{s}^{-1}$
$12\,\mathrm{m}\,\mathrm{s}^{-1}$
$2\,\mathrm{m}\,\mathrm{s}^{-1}$

Solution:

Given Data

Length of pendulum,

l=200\,\mathrm{cm}=2\,\mathrm{m}

Released from horizontal position

\rightarrow

Initial height (relative to mean position)

l=2\,\mathrm{m}

Acceleration due to gravity,

g=10\,\mathrm{m/s}^2

10 \%

of energy lost due to air resistance

Step 1: Initial Potential Energy

At horizontal position (bob just released):

E_{\text{initial}} = m g h = m g l = m(10)(2) = 20\,m\,\mathrm{J}

Step 2: Energy Lost in Air Resistance

10 \%

of energy is lost

\rightarrow

only

90 \%

remains.

E_{\text{mean}} = 0.9 \times E_{\text{initial}} = 0.9 \times 20\,m = 18\,m\,\mathrm{J}

Step 3: Energy at Mean Position

At mean position, the bob has only kinetic energy (potential energy

=0

:

\frac{1}{2} m v^2 = 18\,m

v^2 = 36 \Rightarrow v = 6\,\mathrm{m/s}

Final Answer:

v = 6\,\mathrm{m/s}

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