Question

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displayed and released, oscillates with a period of 0.6 s. What is the weight of the body?

Answer 1

Given, Mass of the spring balance, m = 50 kg,
Length of the scale, y = 20 cm = 0.2 m,
Period of oscillation,T = 0.60 s

We know, F = ky [ from spring force ]
And F = mg [ from Newton's 2nd law, weight = mass × acceleration due to gravity.]

so, mg = ky

k = mg/y

= 50 × 9.8/0.2 N/m

= 2450 N/m

Now, time period in case of spring is given by, $T=2\pi\sqrt{\frac{m}{k}}$

so, T² = 4π²m/k => m = T²k/4π²

Put, T = 0.6 s , k = 2450 Nm

so, m = (0.6)² × 2450/4(3.14)²

m = 22.36 kg

Answer 2

Answer:

Given, Mass of the spring balance, m = 50 kg,

Length of the scale, y = 20 cm = 0.2 m,

Period of oscillation,T = 0.60 s

We know, F = ky [ from spring force ]

And F = mg [ from Newton's 2nd law, weight = mass × acceleration due to gravity.]

so, mg = ky

k = mg/y

= 50 × 9.8/0.2 N/m

= 2450 N/m

Now, time period in case of spring is given by,

so, T² = 4π²m/k => m = T²k/4π²

Put, T = 0.6 s , k = 2450 Nm

so, m = (0.6)² × 2450/4(3.14)²

m = 22.36 kg