Answer to Question #110730 in Mechanics | Relativity for Joey

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?
1
Expert's answer
2020-04-20T19:03:47-0400

k×xmax22=m×vmax22;xmax=m×vmax2k;xmax=0.05×5.65212/5=0.357m\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

Joey
21.04.20, 02:39

Thank you. I ended up doing vmax/ square root m/k which also gave me 0.357m thanks again!

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