Question

A body of mass 5 kg hangs from a spring and oscillates with a time period of $2\pi$ seconds. If the ball is removed, the length of the spring will decrease by

• A. g/k metres
• B. k/g metres
• C. $2 \pi$ meters
• D. g metres

Answer 1

Answer: (D)

g metres

Answer 2

$T=2 \pi sec$
Mass $=5 \,kg$
Spring constant $=k$
$\omega=1 \,rad / sec$
Now $\omega =\sqrt{\frac{k}{m}}$
So, $k=5$
Equilibrium position when it is oscillating is at
$k \Delta x=m g$
or $\Delta x=\frac{m g}{k}$
When the mass is removed the spring will return to its natural length, which is $\Delta x$ upwards.
Since $m=5$ and $k=5$
$\Delta x=g$ metre