Answer to Question #182154 in Mechanical Engineering for luke

Question

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

Answer 1

$m_1 = 2.5kg\\ F_1 = 2.5× 9.81\\ x_1= 5cm = 0.05m$

$k = \dfrac{F_1}{x_1} =\dfrac{2.5×9.81}{0.05}= 490.5N/m$

$m_n = 2kg$

$m_2 = 4.5kg\\ F_2 = 4.5× 10 = 45N\\ x_2 = 6cm = 0.06m$

$k_n =\dfrac{F_1}{x_1} =\dfrac{4.5×9.81}{0.06}= 735.75N/m$

frequency of vibration = $\dfrac{1}{2π}\sqrt{\dfrac{k_n}{m_2}}$

$= \dfrac{1}{2π}\sqrt{\dfrac{735.75}{4.5}}= 2.04 \ Hz$