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**Hint:**First find the spring constant and then by using the equation that gives the time period of oscillation of a spring in relation to mass of the body and the spring constant find the weight of the body. In other words, assume you suspend a body of maximum weight that the scale can read. This would mean that the extension of the spring would be the entire length of the spring. Find the spring constant, and then proceed with using the time period of oscillation to find the mass of the body and subsequently, the weight.

**Formula used:**

Force acting on the body $F_{gravitational} = mg$ where m is the mass of the body and g is the acceleration due to gravity.

Spring force (aka Hooke’s Law) $F_{spring} = ky$, where k is the spring constant and y is the displacement (extension or contraction) of the spring.

Time period of oscillation for a spring $T = 2\pi\sqrt{\dfrac{M}{k}}$, where M is the mass of the body.

**Complete answer:**

Let us first try and understand the parameters of the question.

We have a spring balance that can read a maximum mass of $50\;kg$. Therefore, the maximum force that it can experience is:

$F_{gravitational}= mg = 50 \times 9.8 = 490N$

Under the experience of this force, the spring reacts to the displacement(y) as follows:

$ F_{spring} = ky $

The resistance to extension or the characteristic spring constant is thus given by:

$k =\dfrac{F_{spring}}{y}$ where here $ F_{spring} = F_{gravitational} = 490N$ at maximum extension ($y= 20\;cm = 0.2\;m$).

$\Rightarrow k = \dfrac{490}{0.2} = 2450 Nm^{-1}$

Now, we have a body of unknown weight suspended from the spring, which when released oscillates with a period of $0.6\;s$, i.e.,

$T = 2\pi\sqrt{\dfrac{M}{k}} = 0.6\;s$

Squaring the above equation and re-arranging the relation, we can find the mass of the suspended body:

$\Rightarrow M = \dfrac{T^2 k}{4\pi^{2}} = \dfrac{0.6^2 \times 2450}{4 \times \pi^2} = 22.34\;kg$

Therefore, the weight of the body will be $w =Mg = 22.34 \times 9.8 = 218.95\;N$

**Note:**

Remember that the spring constant remains the same for a given spring within its elastic limit. Though from the equations it may seem like the spring constant is proportional to the force acting on it and inversely proportional to its extension and might hence change with different values of weight and displacement, it is important to understand that the weight and displacement follow a linear relation, i.e., a graph between force and displacement would yield a straight line since successive increments in weight would produce increments in extension by a common factor, which we call as the spring “constant”.

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