Answer to Question #175921 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. How long will it take for the mass to return to its point of release for the second time in seconds?

Answer 1

We can start by finding the spring constant

$k(0.05)=2.5 \times9.81$

$k= \frac{2.5 \times9.81}{(0.05)}=490.5 N/m$

We know time= 2T

$\omega_n= \sqrt{\frac{k}{m}}= \sqrt{\frac{g}{ \triangle x}}= \sqrt{\frac{9.81}{0.06}}=12.78$

$T=\frac{2 \pi}{\omega}$. This implies that $2T=\frac{4 \pi}{\omega}=\frac{4 \pi}{12.78}=0.98 s$