Question

A spring balance has a scale that reads from $0$ to $50\text{kg}$. The length of the scale is $20\text{cm}$. A body suspended from his balance, when displaced and released, oscillates with a period of $0.6\text{s}$. What is the weight of the body ? (Assume that ${\pi }^2=10,g=10{\text{m/s}}^2$)

Answer 1

The scale is $20\text{cm}$ long and reads up to $50\text{kg}$.
i.e., for $50\text{kg}$ displacement is $20\text{cm}$.
$W=Kx;W=50\times 10\text{N}\text{and}x=20\text{cm}=0.20\text{m}$

$K=\frac{W}{x}=\frac{50\times 10}{0.20}=50\times 50=2500\text{N/m}$

Now
$T=2\pi \sqrt{\frac{m}{K}}m=\frac{K{T}^2}{4{\pi }^2}$

Putting values $T=0.6\text{s},K=2500\text{N/m}$ we get
$m=\frac{2500\times {0.6}^2}{4\times 10}=22.5\text{kg}$
Weight of the body $=mg=22.5\times 10\text{N}=225\text{N}$

Why this question? Spring balance is such calibrated that the scale it reads is proportional to the weight of the body suspended from the spring and the proportional constant is the spring constant.