/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A flat square of side \(s_{0}\) ... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 25

A flat square of side \(s_{0}\) at temperature \(T_{0}\) expands by \(\Delta s\) in both length and width when the temperature increases by \(\Delta T\). The original area is \(s_{0}^{2}=A_{0}\) and the final area is $\left(s_{0}+\Delta s\right)^{2}=A .\( Show that if \)\Delta s \ll s_{0}$$$\frac{\Delta A}{A_{0}}=2 \alpha \Delta T$$(Although we derive this relation for a square plate, it applies to a flat area of any shape.)

Short Answer

Expert verified
Answer: The relationship is given by the expression: \(\frac{\Delta A}{A_0} = 2\alpha\Delta T\).

Step by step solution

01

Find the final area A

We are given that the final area is given by: $$A = (s_0 + \Delta s)^2$$
02

Calculate the change in area, \(\Delta A\)

To find the change in area, we subtract the initial area from the final area: $$\Delta A = A - A_0 = (s_0 + \Delta s)^2 - s_0^2$$
03

Apply the binomial approximation

Since \(\Delta s \ll s_0\), we can use the binomial approximation, which states that \((1 + x)^n \approx 1 + nx\) for \(x \ll 1\). In our case: $$\frac{s_0+\Delta s}{s_0} = 1 + \frac{\Delta s}{s_0} \Longrightarrow \left(1 + \frac{\Delta s}{s_0}\right)^2 \approx 1 + 2\frac{\Delta s}{s_0}$$ So, $$A = s_0^2 \left(1 + 2\frac{\Delta s}{s_0}\right)$$
04

Substitute the approximation into the expression for \(\Delta A\)

Using the approximation from Step 3, we can rewrite the expression for \(\Delta A\): $$\Delta A = s_0^2 \left(1 + 2\frac{\Delta s}{s_0}\right) - s_0^2 = 2s_0\Delta s$$
05

Relate \(\Delta s\) to \(s_0\), \(\alpha\), and \(\Delta T\)

The linear expansion coefficient \(\alpha\) is defined as: $$\Delta s = s_0\alpha\Delta T$$ Substituting this into the expression for \(\Delta A\), we get: $$\Delta A = 2s_0^2\alpha\Delta T$$
06

Show the relationship \(\frac{\Delta A}{A_0} = 2\alpha\Delta T\)

We are now ready to write the expression for \(\frac{\Delta A}{A_0}\): $$\frac{\Delta A}{A_0} = \frac{2s_0^2\alpha\Delta T}{s_0^2} = 2\alpha\Delta T$$ This is the relationship between the change in area, the initial area, the linear expansion coefficient, and the change in temperature.

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