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[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

Here we go ~

As we have been given, The spring of spring constant " m " was cut to form 2 new springs in ratio of 5 : 3.

we have to find out the spring constant of the longer spring which was formed, and it's given that spring constant of the main spring was " m "

Now, let the length of main spring be " l "

So, lengths of the resultant springs will be :

[tex]\qquad \cal \dashrightarrow \: \dfrac{5l}{8} \: \: and \: \: \dfrac{3l}{8} \: \: \: \: respectively[/tex]

And we know that spring constant is inversely proportional to length of spring so, we infer that :

[tex]\qquad \cal \dashrightarrow \: m\propto \dfrac{1}{l} m \: \: \: \: \: - (1)[/tex]

where, m is spring constant of main spring and l is its length.

[tex]\qquad \cal\dashrightarrow \: m_1 \propto \dfrac{8}{5l} \: \: \: \: \: \: \: - (2)[/tex]

where, m1 is spring constant of longer spring, and 5l/8 is its length.

let's divide (2) by (1), we will get ~

[tex]\qquad \cal \dashrightarrow \: \dfrac{m_1}{m} = \dfrac{8}{5l} \div \dfrac{1}{l} [/tex]

[tex]\qquad \cal \dashrightarrow \: \dfrac{m_1}{m} = \dfrac{8}{5l} \times \dfrac{l}{1} [/tex]

[tex]\qquad \cal \dashrightarrow \: \dfrac{m_1}{m} = \dfrac{8}{5} [/tex]

[tex]\qquad \cal \dashrightarrow \: {m_1}{} = \dfrac{8}{5} m [/tex]

So, the spring constant of equivalent longer strong formed will be 8/5 m

Ankit

prove:

[tex]m_2 = \frac{8m}{5} [/tex]

Explanation:

Given:

Spring constant of spring= m

The ratio of spring = 5:3

To find:

spring constant of the larger piece of spring?

Solution:

Let the length of larger piece L2 be 5x and smaller piece L1 be 3x with spring constant m2 and m1 respectively.

now,

spring constant (m) inversely proportional to the length of spring i.e.

[tex]m \propto \: \frac{1}{l} {,}\: m_1 \propto \: \frac{1}{l_1} {,}\: m_2 \propto \: \frac{1}{l_2}[/tex]

Also,

The spring is connected in series hence,

[tex] \frac{1}{m} = \frac{1}{m_1} + \frac{1}{m_2} [/tex]

and

[tex] \frac{m_1}{m_2} = \frac{l_2}{l_1} \\ {m_1} = {m_2} \cdot \frac{l_2}{l_1} \\ {m_1} = {m_2} \cdot \frac{5}{3} \\ [/tex]

Substituting above value in,

[tex] \frac{1}{m} = \frac{1}{m_1} + \frac{1}{m_2} \\ \frac{1}{m} = \frac{1}{{m_2} \cdot \frac{5}{3}} + \frac{1}{m_2} \\ \frac{1}{m} = \frac{3}{5m_2} + \frac{1}{m_2} \\ \frac{1}{m} = \frac{3m_2 + 5m_2}{5 {m_2}^{2} } \\ \frac{1}{m} = \frac{8m_2}{5 {m_2}^{2} } \\ \frac{1}{m} = \frac{8 \cancel {m_2}}{5 \cancel{m_2}^{2} } \\ \frac{1}{m} = \frac{8}{5m_2} \\ m_2 = \frac{8m}{5} [/tex]

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