Answer to Question #182154 in Mechanical Engineering for luke

Question #182154

A spring extends by 5 centimeters when loaded vertically with 2.5kg of weight. A new weight of 2kg was attached to it and was displaced 6cm from its new equilibrium position and was released from rest. Determine the frequency of vibration in Hz


1
Expert's answer
2021-04-21T07:05:49-0400

"m_1 = 2.5kg\\\\\nF_1 = 2.5\u00d7 9.81\\\\\nx_1= 5cm = 0.05m"


"k = \\dfrac{F_1}{x_1} =\\dfrac{2.5\u00d79.81}{0.05}= 490.5N\/m"



"m_n = 2kg"


"m_2 = 4.5kg\\\\\nF_2 = 4.5\u00d7 10 = 45N\\\\\nx_2 = 6cm = 0.06m"


"k_n =\\dfrac{F_1}{x_1} =\\dfrac{4.5\u00d79.81}{0.06}= 735.75N\/m"



frequency of vibration = "\\dfrac{1}{2\u03c0}\\sqrt{\\dfrac{k_n}{m_2}}"


"= \\dfrac{1}{2\u03c0}\\sqrt{\\dfrac{735.75}{4.5}}= 2.04 \\ Hz"

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