A force of 6.4 N stretches a vertical spring by 0.1 m . mass that must be suspended &nbsp

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13.Oscillations

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A force of $6.4\ N$ stretches a vertical spring by $0.1\ m$. The mass that must be suspended from the spring so that it oscillates with a time period of $\pi/4\ second$ is .... $kg$

A

$\frac{\pi }{4}$

B

$\frac{4 }{\pi}$

C

$1$

D

$10$

Solution

Spring constant $\mathrm{K}=\frac{6.4}{0.1}=64 \mathrm{N} / \mathrm{m}$

Now $\quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

or $\frac{\pi}{4}=2 \pi \sqrt{\frac{\mathrm{m}}{64}}$

$\therefore m=1 \mathrm{kg}$

Standard 11

Physics

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