Question

A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.2 Hz. How far was the block pulled back before being released?

Answer 1

Given frequency , $f=1.2Hz$

so Time Period, $T=1/f=1/1.2=0.83s$

Find spring constant k

$T=2\pi \sqrt{m/k}$

$T^{2}=4\pi^{2}* m/k$

$k=4\pi^{2}* m/T^{2}$

$k=4\pi^{2}* 0.26kg/(0.83s)^{2}$

$k=14.781N/m$

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Given that Force ,F =1.2N

By Hooke's Law

$F=kx$

$x=F/k$

$x=1.2N/(14.781N/m)$

$x=0.0812m$

ANSWER: $x=8.12cm$

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