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Oscillations Part 1

Section: Physics

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Question : 35 of 100

Marks: +1, -0

The period of oscillation of a simple pendulum is

T=2\pi\sqrt{\;\frac{L}{g}}.

Measured value of

L

1.0\ \text{m}

from metre scale having a minimum division of

1\ \text{mm}

and time of one complete oscillation is

1.95\ \text{s}

measured from stopwatch of

0.01\ \text{s}

resolution. The percentage error in the determination of

g

[24 Feb 2021 Shift 2]

$1.13\%$
$1.03\%$
$1.33\%$
$1.30\%$

Solution:

Given,

T=2\pi\sqrt{\;\frac{L}{g}}

. . .

(i)

where, time period,

T=1.95\ \text{s}

Length of string,

I=1\ \text{m}

Acceleration due to gravity

=g

Error in time period,

\Delta T=0.01\ \text{s}=10^{-2}\ \text{s}

Error in length,

\Delta L=1\ \text{mm}=1\times 10^{-3}\ \text{m}

Squaring Eq. (i) on both sides, we get

T^{2}\;\;=4\pi^{2}\;\frac{L}{g}

g\;\;=4\pi^{2}\;\frac{L}{T^{2}}

\Rightarrow\;\; \;\frac{\Delta g}{g}\;\; =\;\frac{\Delta L}{L}+\;\frac{2\Delta T}{T}=\;\frac{10^{-3}}{1}+\;\frac{2\times10^{-2}}{1.95}

\;\; =10^{-3}+1.025\times10^{-2}

\;\; =10^{-3}+10.25\times10^{-3}

\;\; =11.25\times10^{-3}

\because\;\; \frac{\Delta g}{g}\times100\;\; =11.25\times10^{-3}\times10^{2}

\;\; =1.125\%=1.13\%

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