Q. The scale of a spring balance reading from 0 to $10\ kg$ is $0.25\ m$ long. A body suspended from the balance oscillates vertically with a period of $\pi/10$ second. The mass suspended is (neglect the mass of the spring)
A) $10\ kg$
B) $0.98\ kg$
C) $5\ kg$
D) $20\ kg$
Explanation
(b) Using $F = kx\ \Rightarrow 10g = k \times 0.25 \Rightarrow k = \frac{10g}{0.25} = 98 \times 4$
Now $T = 2\pi\sqrt{\frac{m}{k}} \Rightarrow m = \frac{T^{2}}{4\pi^{2}}k$
$\Rightarrow m = \frac{\pi^{2}}{100} \times \frac{1}{4\pi^{2}} \times 98 \times 4 = 0.98\ kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.