Tag: speed of a travelling wave

Velocity of sound in a gas is

$v _{air}=\sqrt{\dfrac{\gamma RT}{M}}=330m/s$

$\displaystyle v= \sqrt {\frac {\gamma RT}{M}}$

$PV=RT$

$\displaystyle P\frac {M}{d}=RT$

$\displaystyle \frac {P}{d} = \frac {RT}{M}$

$\displaystyle v= \sqrt {\frac {\gamma P}{d}}$

$No$

Velocity of sound wave in a medium is given by
$v= \sqrt{\frac{K}{\rho}}$ where K is the bulk modulus and $\rho$ is the density.
The classification of sound waves based on wavelength($\lambda$) is independent of speed of sound in the medium( speed depends on properties of a medium).
Hence, $V _u$, $V _i$ and $V _a$ are all equal.

Speed of sound in stretched string
$v = \dfrac {\overline {T}}{\mu} ..... (i)$
where $T$ is the tension in the string and $\mu$ is mass per unit length.
According to Hooke's law, $F\propto X$
$\therefore T\propto X$ .... (ii)
From Eqs. (i) and (ii)
$v\propto$
$\therefore v' = \overline {1.5V} = 1.22\ V$.

$1$ mole of any gas occupies $22.4$ liters at STP.
Therefore, the density of air at STP is
$\rho = \dfrac{\text{Mass of one mole of air}}{\text{Volume of one mole of air at STP}}$

$= \dfrac{29 \times 10^{-3} \,\,kg}{22.4 \times 10^{-3} \,\,m^3} = 1.29 \,\,kg \,\,m^{-3}$

At STP, $P = 1\,\,atm = 1.01 \times 10^5 \,\,N \,\,m^{-2}$

For velocity of sound in gas
$\displaystyle v=\sqrt {\frac {p\gamma }{p}}$

$[P$ is pressure and $p$ is density of gas, $\gamma$ is $C _p/C _v]$

Here, $\displaystyle v _1 = \sqrt {\frac {\gamma P}{p _1}}$ and $v _2 = \sqrt {\frac {\gamma P}{p _2}}$

$\displaystyle \frac {v _1}{v _2} = \sqrt {\frac {p _2}{p _1}} = \sqrt {\frac {1}{4}}=\frac {1}{2}$

$V _{sound}=\sqrt{\dfrac{\gamma RT}{M}}$ and $V _{rms}=\sqrt{\dfrac{3RT}{M}}$
$\Rightarrow\dfrac{V _{sound}}{V _{rms}}=\sqrt{\dfrac{\gamma}{3}}$
Hence (A) is correct.