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2) A mass-on-a-spring oscillates with an amplitude of 45.0 cm. It has a maximum acceleration of 2.84 ms at the end points of

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Answer #1

For an oscillation, acceleration is

\\a = - \omega^2 x

where \omega is the angular frequency and T is the displacement

and

\\\omega = \frac{2 \pi}{T}......where t is the period

Given, maximum displacement is amplitude(at the ends of motion)

A = 45 \ cm = 0.45 \ m

Given, maximum acceleration is a_{max} = 2.84 \ m/s^2

we have

a_{max} = \omega^2 A

\\\omega^2 = \frac{a_{max}}{A} \\\\\omega^2 = \frac{2.84}{0.45} \\\\\omega^2 = 6.31 \\\\\omega = \sqrt{6.31} \\\\\omega = 2.512 \ rad/s

that is,

\\\frac{2 \pi}{T} = 2.512 \\\\T = \frac{2 \pi}{2.512} \\\\T = 2.501 \ s ......period of oscillation

Given, mass is \\m = 2 \ kg

Let  k  be the spring constant,

We have,

\\\omega = \sqrt{\frac{k}{m}} \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2 \\\\k = 2 * 2.512 \\\\k = 5.024 \ N/m .......spring constant

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