Question

A body of mass $4.9kg$ hangs from a spring and oscillates with a period $0.5s$. On the removal of the body, the spring is shortened by (Take $g=10\text{ m}{s}^{-2},n^2=10$ )

• A. 6.3 m
• B. 0.63 m
• C. 6.25 cm
• D. 63 cm
• E. 0.625 cm

Answer 1

Answer: (C)

Time period of oscillation, $T=2\pi \sqrt{\frac{m}{k}}$ where $m$ is the mass of the body suspended from a spring and $k$ is spring constant of the spring.
or $k=\frac{4{\pi }^{2}m}{T^2}$
Substituting the given values, we get
$k=\frac{4\times 10\times 4.9}{(0.5{)}^2}N{m}^{-1}$
On the removal of the body, the spring is shortened by $x$.
$\therefore mg=kx$
or $x=\frac{mg}{k}=\frac{4.9\times 10\times (0.5{)}^2}{4\times 10\times 4.9}$ (Using (i))
$=\frac{0.25}{4}=0.0625m=6.25cm$