The vertical extension in a light spring by a weight of $1 k g$ suspended from the wire is $9.8 c m$ . The period of oscillation:

20 π s e c

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2 π s e c

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\frac{2 π}{10} s e c

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200 π s e c

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Solution

The correct option is B $\frac{2 π}{10} s e c$
$∵ m g = k x \Rightarrow \frac{m}{k} = \frac{x}{g}$

$\Rightarrow T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{x}{g}}$

$T = 2 π \sqrt{\frac{9.8 \times 10^{- 2}}{9.8}} = \frac{2 π}{10} sec$

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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:

The correct option is B 2π 10sec∵mg=kx⇒mk=xg⇒T=2π√mk=2π√xgT=2π√9.8×10−29.8=2π10sec

The vertical extension in a light spring by a weight of $1 k g$ suspended from the wire is $9.8 c m$ . The period of oscillation:

The correct option is B $\frac{2 π}{10} s e c$
$∵ m g = k x \Rightarrow \frac{m}{k} = \frac{x}{g}$

$\Rightarrow T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{x}{g}}$

$T = 2 π \sqrt{\frac{9.8 \times 10^{- 2}}{9.8}} = \frac{2 π}{10} sec$