# Answer to Question #175918 in Mechanical Engineering for luke

Question #175918

A spring extends by 5 centimeters when loaded vertically with 2.5kg of weight. A new weight of 1.75kg 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
2021-04-01T01:11:18-0400

As per given condition, there are two forces attached to the spring ,

Initial force=(m1)×g

Initial displacement in the spring= y1

Final force= (m1+m2)×g

Final displacement=y1+y2

Displacement (y1) =5cm=0.05 m

Mass attached (m1)=2.5 kg

Displacement (y2) =6cm=0.06 m

Mass attached (m2)=1.75 kg

If k is the spring constant, It is the ratio of change in load to change in displacement,

"k= \\frac{m_2 \u00d7g}{y_2}"

We know that ,

"\\frac{k}{m_2}=\\frac{g}{y_2}=163.33 s^{-1}"

The frequency of spring is given by,

"f= \\sqrt{\\frac{k_2}{m_2}}=\\sqrt{163.33}=12.78 \\space s^{-1}"

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