A $100 g$ mass stretches a particular spring by $9.8 c m,$ when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28 s$ ?

1000 g

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10^{5} g

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10^{7} g

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10^{4} g

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Solution

The correct option is C $10^{4} g$
$\begin{matrix} m = 0.1 k g, x = 9.8 \times 10^{- 2} m, T = 6.28 s K = \frac{m g}{x} \Rightarrow k = 10 T = 2 π \sqrt{\frac{M}{K}} \Rightarrow 6.28 = 2 \times 3.14 \sqrt{\frac{M}{10}} 1 = \frac{M}{10} \Rightarrow M = 10 k g = 10^{4} g \end{matrix}$

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A 100g mass stretches a particular spring by 9.8 cm, when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be 6.28 s?

The correct option is C 104 gm=0.1kg,x=9.8×10−2m,T=6.28sK=mgx⇒k=10T=2π√MK⇒6.28=2×3.14√M101=M10⇒M=10kg=104g

A $100 g$ mass stretches a particular spring by $9.8 c m,$ when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28 s$ ?

The correct option is C $10^{4} g$
$\begin{matrix} m = 0.1 k g, x = 9.8 \times 10^{- 2} m, T = 6.28 s K = \frac{m g}{x} \Rightarrow k = 10 T = 2 π \sqrt{\frac{M}{K}} \Rightarrow 6.28 = 2 \times 3.14 \sqrt{\frac{M}{10}} 1 = \frac{M}{10} \Rightarrow M = 10 k g = 10^{4} g \end{matrix}$