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 Hz

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,

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