what is the answer? Print calculator Periodic Table Question 8 of 11 Map A A Sap

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what is the answer?

Print calculator Periodic Table Question 8 of 11 Map A A Sapling Learning An undamped 1.98 kg horizontal spring oscillator has a spring constant of 33.2 N/m. While oscillating, it is found to have a speed of 3.16 m/s it passes through its equilibrium position. What is its amplitude of Number What is the oscillator's total mechanical energy as it passes through a position that is 0.738 of the amplitude away from the equilibrium position? click to edit Number Ava Due Poir Gra Polic You You C up or You C you g There O eTex O Help Web

Explanation / Answer

, the angular frequency for the spring = (k/m) = [(33.2N/m)/1.98kg] = 4.09rad/s

At equilibrium position the speed is v = A

so 3.16m/s = A = A(4.09rad/s)

or Amplitude A = 0.772m

Total mechanical enery is given by E = (1/2)m2A2 = (1/2)(1.98kg)(4.09rad/s)2(0.772m)2

or E = 9.87J

We can see that the total mechanical energy is independent of position of oscillator. So the total mechanical energy would be E = 9.87J

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige.....

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