Question Number 43694 by Tawa1 last updated on 14/Sep/18
$$\mathrm{Using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{dimension},\:\mathrm{derive}\:\mathrm{an}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{the}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{sound}\:\mathrm{waves}\:\left(\mathrm{v}\right)\:\mathrm{through}\:\mathrm{a}\:\mathrm{medium}.\:\mathrm{Assume}\:\mathrm{that}\:\mathrm{the}\:\mathrm{velocity}\: \\ $$$$\mathrm{depends}\:\mathrm{on}:\:\:\left(\mathrm{i}\right)\:\mathrm{Modulus}\:\mathrm{of}\:\mathrm{elasticity}\:\left(\mathrm{E}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{medium} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{density}\:\mathrm{of}\:\mathrm{the}\:\mathrm{medium}\:\left(\rho\right),\:\mathrm{take}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{K}\:=\:\mathrm{1} \\ $$
Answered by alex041103 last updated on 14/Sep/18
![We know that E=[(N/m^2 )]=[((kg(m/s^2 ))/m^2 )]=[m^(−1) s^(−2) kg^1 ] ρ=[((kg)/m^3 )]=[m^(−3) kg^1 ] v=[(m/s)]=[ms^(−1) ] ⇒We suppose v=K E^α ρ^β ⇒ms^(−1) =m^(−α) s^(−2α) kg^α m^(−3β) kg^β ⇒m^1 s^(−1) kg^0 =m^(−(α+3β)) s^(−2α) kg^(α+β) −α−3β=1⇒α+3β=−1 −2α=−1⇒α=(1/2) α+β=0⇒β=−α=−(1/2) And α+3β=(1/2)+3(−(1/2))=−1 ⇒α=1/2 and β=−1/2 ⇒v=K E^(1/2) ρ^(−1/2) But K=1 ⇒v=(√(E/ρ))](Q43701.png)
$$ \\ $$$$ \\ $$$${We}\:{know}\:{that} \\ $$$${E}=\left[\frac{{N}}{{m}^{\mathrm{2}} }\right]=\left[\frac{{kg}\frac{{m}}{{s}^{\mathrm{2}} }}{{m}^{\mathrm{2}} }\right]=\left[{m}^{−\mathrm{1}} {s}^{−\mathrm{2}} {kg}^{\mathrm{1}} \right] \\ $$$$\rho=\left[\frac{{kg}}{{m}^{\mathrm{3}} }\right]=\left[{m}^{−\mathrm{3}} {kg}^{\mathrm{1}} \right] \\ $$$${v}=\left[\frac{{m}}{{s}}\right]=\left[{ms}^{−\mathrm{1}} \right] \\ $$$$\Rightarrow{We}\:{suppose}\:{v}={K}\:{E}^{\alpha} \rho^{\beta} \\ $$$$\Rightarrow{ms}^{−\mathrm{1}} ={m}^{−\alpha} {s}^{−\mathrm{2}\alpha} {kg}^{\alpha} \:{m}^{−\mathrm{3}\beta} {kg}^{\beta} \: \\ $$$$\Rightarrow{m}^{\mathrm{1}} {s}^{−\mathrm{1}} {kg}^{\mathrm{0}} ={m}^{−\left(\alpha+\mathrm{3}\beta\right)} {s}^{−\mathrm{2}\alpha} {kg}^{\alpha+\beta} \\ $$$$ \\ $$$$−\alpha−\mathrm{3}\beta=\mathrm{1}\Rightarrow\alpha+\mathrm{3}\beta=−\mathrm{1} \\ $$$$−\mathrm{2}\alpha=−\mathrm{1}\Rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\alpha+\beta=\mathrm{0}\Rightarrow\beta=−\alpha=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${And}\:\alpha+\mathrm{3}\beta=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{1} \\ $$$$\Rightarrow\alpha=\mathrm{1}/\mathrm{2}\:{and}\:\beta=−\mathrm{1}/\mathrm{2} \\ $$$$\Rightarrow{v}={K}\:{E}^{\mathrm{1}/\mathrm{2}} \rho^{−\mathrm{1}/\mathrm{2}} \\ $$$${But}\:{K}=\mathrm{1} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{E}}{\rho}} \\ $$
Commented by Tawa1 last updated on 14/Sep/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by alex041103 last updated on 14/Sep/18
$${You}\:{are}\:{welcome}! \\ $$