Question

The length of scale of a spring balance which reads to $100kg$ is $20cm$.On suspending a packet from spring it oscillates with a frequency of $5$ oscillation second in a vertical plane. If $g={\pi }^{2}m/se{c}^2$. Calculate the weight of the suspended object.

Answer 1

Maximum mass read by the scale =M= 100 kg

maximum stretch = l = 20 cm

Maximum force that can be exerted on the spring = F = Mg = $100kg\times 9.8m/s^2=980kgm/s^2=980N$

Spring constant = $k=\frac{F}{l}=\frac{980N}{0.2m}=4900N/m$

Time period of oscillation is given as:$T=\frac{1}{\nu }=\frac{1}{5}s=2\pi \sqrt{\frac{m}{k}}$

where, m = mass of the object suspended

$m=\frac{T^{2}k}{4{\pi }^2}$

$m=\frac{\left(1/25\right)\times 4900)}{4\times 3.14\times 3.14}=4.97kg$

So, weight of the object suspended = $4.97\times 9.8m/s^2=48.7N$