Mechanical Properties of Solids

If $\mu$  is Poisson's ratio,

Y = 3K (1 - 2 $\mu$ )      ……… (1)

and Y = 2 $n\left(1+\mu \right)$  ……… (2)

With the help of equations (1) and (2), we can write

  $\frac{3}{Y}=\frac{1}{\eta }+\frac{1}{3k}\Rightarrow K=\frac{\eta Y}{9\eta -3Y}$

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 * (0.5*10? * 10? / 0.1) * (0.04)² = 20*10? ³ v²
0.5 * (5*10²) * 1.6*10? ³ = 20*10? ³ v²
0.4 = 20*10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s

As we know that

$Y=\frac{FL}{A?L}$          

$?L=0.04m=\frac{FL}{AY}...............\left(i\right)$           

If length and diameter both are doubled

$?L\text{'}=\frac{F.2L}{4A.Y}=\frac{FL}{2Y}=0.02m=2cm$       

Energy density (u) = $\frac{1}{2}*\sigma *\epsilon =\frac{1}{2}*Y\epsilon *\epsilon =\frac{1}{2}{\epsilon }^{2}Y$ $\frac{}{}\frac{}{}\frac{}{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}{\epsilon }^{2}YA$

Strain $\left(\epsilon \right)=\frac{\Delta \mathcal{l}}{{\mathcal{l}}_0}=\alpha \Delta T=10^{-5}*10=10^{-4}$ $\frac{}{{}_{}}{}^{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}*10^{-8}*10^{11}*10^{-2}=5$

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T0 (developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ $$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$

At steady state :-

$\frac{F}{A/2}=Y\frac{\Delta \mathcal{l}}{\mathcal{l}}\Rightarrow F=\frac{Y\Delta \mathcal{l}A}{2\mathcal{l}}$

$=\frac{YA}{2}=\frac{2*10^{11}*10^{-4}}{2}=10^{7}N$

Slope = $\frac{\Delta L/W}{L}=\frac{\Delta L/L}{W}=\frac{1}{YA}$ $\frac{}{}\frac{}{}\frac{}{}$

$Y=\frac{1}{\left(Slope\right)A}$

$Y=\frac{1}{\left(2*10^{-6}\right)\left(0.25*10^{-5}\right)}$

$Y=2*10^{11}N/m^2$

$KE=\Delta U$

$\Delta U=\frac{1}{2}m{v}^2$

$n{C}_v\Delta T=\frac{m{v}^2}{2}$

$\Rightarrow \frac{m}{M}\frac{R}{y-1}\Delta T=\frac{m{v}^2}{2}$

$\Delta T=\frac{0.4}{2}\frac{M{v}^2}{R}$

$\Delta T=\frac{M{v}^2}{5R}$