A force of 6.4 , N stretches a vertical sprin | vedclass

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(B) The force constant $k$ of the spring is determined by Hooke's Law: $F = kx$.Given $F = 6.4 \, N$ and $x = 0.1 \, m$,we have $6.4 = k(0.1)$,which g

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(B) The force constant $k$ of the spring is determined by Hooke's Law: $F = kx$.Given $F = 6.4 \, N$ and $x = 0.1 \, m$,we have $6.4 = k(0.1)$,which gives $k = 64 \, N/m$.The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Substituting the given values: $\frac{\pi}{4} = 2\pi \sqrt{\frac{m}{64}}$.Dividing both sides by $\pi$: $\frac{1}{4} = 2 \sqrt{\frac{m}{64}}$.Dividing by $2$: $\frac{1}{8} = \sqrt{\frac{m}{64}}$.Squaring both sides: $\frac{1}{64} = \frac{m}{64}$.Therefore,$m = 1 \, kg$.

$A$ 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 period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

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