Question #110730

A spring in simple harmonic motion is found to have a velocity of 5.65 m/s when a mass of 50.0 grams is placed on the end of the spring. If the spring constant is 12.5 N/m, what is the maximum

displacement from equilibrium position that the spring system experiences?

displacement from equilibrium position that the spring system experiences?

Expert's answer

"\\frac{k\\times x_{max}^2}{2}=\\frac{m\\times v_{max}^2}{2};\\\\x_{max}=\\sqrt{\\frac{m\\times v_{max}^2}{k}};\\\\x_{max}=\\sqrt{\\frac{0.05\\times 5.65^2}{12\/5}} =0.357m"

Answer: 0.357m

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## Comments

Joey21.04.20, 02:39Thank you. I ended up doing vmax/ square root m/k which also gave me 0.357m thanks again!

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