Question

A vertical spring carries a $5kg$ body and is hanging in equilibrium, an additional force is applied so that the spring is further stretched. When released from this position, it performs $50$ complete oscillation is $25s$, with an amplitude of $5cm$. The additional force applied is -

• A. $80N$
• B. $80{\pi }^{2}N$
• C. $4{\pi }^{2}N$
• D. $4N$

Answer 1

Answer: (C)

$4{\pi }^{2}N$

$T=\frac{25}{50}=\frac{1}{2}s\Rightarrow \omega =\frac{2\pi }{T}=4\pi rad/s$
Spring constant $k=m{\omega }^2=5\times (4\pi {)}^2=80{\pi }^{2}N/m$
Force required o stretch the spring by $5cm$ is
$F=kx=80{\pi }^2\times 0.05N=4{\pi }^2$