Answer to Question #175919 in Mechanical Engineering for luke

Question

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 rad per second.

Answer 1

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,

$k= \frac{m_2 ×g}{y_2}$

We know that ,

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

The frequency of spring is given by,

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