Vertical extension in a light spring by a weight of 1, kg suspended from wire is 9.8, cm . period of

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13.Oscillations

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The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. The period of oscillation

A

$20\pi \sec $

B

$2\pi \sec $

C

$2\pi /10\sec $

D

$200\pi \sec $

Solution

(c) $mg = kx$

==> $\frac{m}{k} = \frac{x}{g}$

==> $T = 2\pi \sqrt {\frac{m}{k}} = 2\pi \sqrt {\frac{x}{g}} $

$ = 2\pi \sqrt {\frac{{9.8 \times {{10}^{ – 2}}}}{{9.8}}} = \frac{{2\pi }}{{10}}\,sec$

Standard 11

Physics

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