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Citrus2011 [14]

3 years ago

12

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

aced from its equilibrium position and undergoes simple harmonic oscillations. What is the period of the oscillations

1 answer:

pantera1 [17]3 years ago

7 0

Answer:

The period is $T = 0.700 \ s$

Explanation:

From the question we are told that

The mass is $m = 0.350 \ kg$

The extension of the spring is $x = 12.0 \ cm = 0.12 \ m$

The spring constant for this is mathematically represented as

$k = \frac{F}{x}$

Where F is the force on the spring which is mathematically evaluated as

$F = mg = 0.350 * 9.8$

$F =3.43 \ N$

So

$k = \frac{3.43 }{ 0.12}$

$k = 28.583 \ N/m$

The period of oscillation is mathematically evaluated as

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

substituting values

$T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }$

$T = 0.700 \ s$

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Explanation:

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Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

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Knowing this area we can isolate $l_{o}$ from (1):

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And substitute $l_{o}$ in (2):

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$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

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