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Home » A spring balance has a scale that reads from 0 to . The length of the scale is A body suspended from this balance, when displaced and released, oscillates with a period of S. What is the weight of the body?
A spring balance has a scale that reads from 0 to 50 \mathrm{~kg}. The length of the scale is 20 \mathrm{~cm} . A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 S. What is the weight of the body?
CBSE | Class 11 | NCERT | Oscillations | Physics
A spring balance has a scale that reads from 0 to 50 \mathrm{~kg}. The length of the scale is 20 \mathrm{~cm} . A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 S. What is the weight of the body?

Maximum mass that the scale can rea is given as M=50 \mathrm{~kg}

Maximum displacement of the spring = Length of the scale, I=20 \mathrm{~cm} =0.2 \mathrm{~m}

Time period is given as T=0.6 \mathrm{~s}

Maximum force exerted on the spring can be represented as

\mathrm{F}=\mathrm{mg}

where,

\begin{array}{l} \mathrm{g}=\text { acceleration due to gravity }=9.8 \mathrm{~m} / \mathrm{s}^{2} \\ \mathrm{~F}=50 \times 9.8=490 \end{array}

So,

Spring constant can be calculated as \mathrm{k}=\mathrm{F} / \mathrm{l}

=490 / 0.2

We get,

=2450 \mathrm{~N} \mathrm{~m}^{-1}

Mass \mathrm{m} is suspended from the balance.

Time period will be \mathrm{t}=2 \pi \sqrt{\mathrm{m}} / \mathrm{k}

So,

\begin{array}{l} \mathrm{m}=(\mathrm{T} / 2 \pi)^{2} \mathrm{x} \mathrm{k} \\ =\{0.6 /(2 \times 3.14)\}^{2} \times 2450 \end{array}

We get,

=22.36 \mathrm{~kg}

Hence, weight of the body will be =m g=22.36 \times 9.8

=219.13 \mathrm{~N}

As a result, the weight of the body is about 219 \mathrm{~N}

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