Question

A spring balance has a scale that reads from $0$ to $50kg$. The length of the scale is $20cm$. A block of mass $m$ is suspended from this balance, when displaced and released, it oscillates with a period $0.5s$. The value of $m$ is (Take $g=10m{s}^{-2})$

• A. $8kg$
• B. $12kg$
• C. $16kg$
• D. $20kg$

Answer 1

Answer: (C)

The $20cm$ length of the scaler reads upto $50kg$.
$\therefore F=mg=(50kg)(10m{s}^{-2})=500N$ and $x=20cm=0.2m$
$\therefore$ Spring constant, $k=\frac{F}{x}=\frac{500N}{0.2m}$
$=2500N{m}^{-1}$
As $T=2\pi \sqrt{\frac{m}{k}}$
Squaring both sides, we get
$T^2=\frac{4{\pi }^{2}m}{k}$
$m=\frac{T^{2}k}{4{\pi }^2}$
$=\frac{{\left(0.5s\right)}^2\times \left(2500N{m}^{-1}\right)}{4\times {\left(3.14\right)}^2}$
$=16kg$