A 2.20 kg mass is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 22.7 N is required to hold the mass at rest when it is pulled 0.205 m from itsequilibrium position (the origin of the x axis). The mass is now released from rest with an initial displacement of xi = 0.205 m, and it subsequently undergoes simple harmonic oscillations. Calculate the speed when the displacement equals one third of the maximum value.Incorrect. Tries 4/6 Previous TriesSubmit AnswerCalculate the acceleration when the displacement equals one third of the maximum value.Tries 0/6Submit Answer

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Asked Nov 21, 2019
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A 2.20 kg mass is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 22.7 N is required to hold the mass at rest when it is pulled 0.205 m from its
equilibrium position (the origin of the x axis). The mass is now released from rest with an initial displacement of xi = 0.205 m, and it subsequently undergoes simple harmonic oscillations.
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A 2.20 kg mass is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 22.7 N is required to hold the mass at rest when it is pulled 0.205 m from its equilibrium position (the origin of the x axis). The mass is now released from rest with an initial displacement of xi = 0.205 m, and it subsequently undergoes simple harmonic oscillations.

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Calculate the speed when the displacement equals one third of the maximum value.
Incorrect. Tries 4/6 Previous Tries
Submit Answer
Calculate the acceleration when the displacement equals one third of the maximum value.
Tries 0/6
Submit Answer
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Calculate the speed when the displacement equals one third of the maximum value. Incorrect. Tries 4/6 Previous Tries Submit Answer Calculate the acceleration when the displacement equals one third of the maximum value. Tries 0/6 Submit Answer

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Expert Answer

Step 1

The spring constant of the spring is

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F k = Fis the horizontal force and x is the distance moved from its equlibrium position 22.7N 0.205m =110.73N/m

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Step 2

The angular velocity is

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|m 110.73N/m 2.20kg = 7.09 rad/s

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Step 3

(a)

The speed when the displacement equals...

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A2-(1/3)) 7.09rad/s(0.205m)-((1/3)(0.205) =1.37 m/s

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