Sound Waves (Longitudinal wave) and it’s Characteristics (Speed etc.) Questions in English

(C) For an adiabatic process,the relation between temperature $T$ and pressure $p$ is given by $T^\gamma p^{1-\gamma} = \text{constant}$.
Taking the natural logarithm on both sides: $\gamma \ln T + (1-\gamma) \ln p = \text{constant}$.
Differentiating both sides: $\gamma \frac{\Delta T}{T} + (1-\gamma) \frac{\Delta p}{p} = 0$.
Rearranging for $\Delta T$: $\frac{\Delta T}{T} = \frac{\gamma - 1}{\gamma} \frac{\Delta p}{p}$.
Given: $T = 273 \ K$ (at $NTP$),$\gamma = 1.5$,$\Delta p = 0.001 \ dyne/cm^2$,$p = 1.013 \times 10^6 \ dyne/cm^2$.
Substituting the values: $\Delta T = 273 \times \left( \frac{1.5 - 1}{1.5} \right) \times \frac{0.001}{1.013 \times 10^6}$.
$\Delta T = 273 \times \frac{0.5}{1.5} \times \frac{10^{-3}}{1.013 \times 10^6} = 273 \times \frac{1}{3} \times 0.987 \times 10^{-9} \approx 8.97 \times 10^{-8} \ K$.

(C) The speed of the car is $v_c = 72 \,km/h = 72 \times \frac{5}{18} = 20 \,m/s$.
In $t = 10 \,s$,the distance traveled by the car is $d_c = v_c \times t = 20 \times 10 = 200 \,m$.
Initially,the car is at $1800 \,m$ from the hill. After $10 \,s$,the car has moved $200 \,m$ closer to the hill,so its new distance from the hill is $1800 - 200 = 1600 \,m$.
The sound travels from the car to the hill $(1800 \,m)$ and then reflects back to the car at its new position $(1600 \,m)$.
Total distance traveled by sound $D = 1800 \,m + 1600 \,m = 3400 \,m$.
Since the time taken is $10 \,s$,the speed of sound $v_s = \frac{D}{t} = \frac{3400}{10} = 340 \,m/s$.

(C) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma R T}{M}}$.
From this relation,it is clear that the speed of sound $v$ is independent of pressure $p$ and is directly proportional to the square root of the absolute temperature $T$,i.e.,$v \propto \sqrt{T}$.
Given initial conditions: $T_1 = T$ and $v_1 = v$.
Given final conditions: $T_2 = 2 T$ and $p_2 = \frac{p}{2}$.
Since $v$ is independent of pressure,the change in pressure does not affect the speed.
Using the proportionality $v \propto \sqrt{T}$,we have:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
$\frac{v_2}{v} = \sqrt{\frac{2 T}{T}} = \sqrt{2}$
Therefore,$v_2 = \sqrt{2} v$.

(C) Given: Path difference $\Delta x = 50 \ cm = 0.5 \ m$,Phase difference $\Delta \phi = 1.8 \pi$,Speed of sound $v = 340 \ ms^{-1}$.
We know the relation between path difference and phase difference is $\Delta x = \frac{\lambda}{2 \pi} \cdot \Delta \phi$.
Substituting the values: $0.5 = \frac{\lambda}{2 \pi} \times 1.8 \pi$.
$0.5 = \lambda \times 0.9 \Rightarrow \lambda = \frac{0.5}{0.9} = \frac{5}{9} \ m$.
The frequency $f$ is given by $f = \frac{v}{\lambda}$.
$f = \frac{340}{5/9} = \frac{340 \times 9}{5} = 68 \times 9 = 612 \ Hz$.

(C) The frequency of a sound wave is a characteristic property of the source that produces it.
When a wave travels from one medium to another (e.g.,from air to water),its speed and wavelength change,but its frequency remains constant.
Therefore,the frequency of the wave at the bottom of the reservoir remains $v \text{ Hz}$.

(C) The velocity of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma R T}{M}} = \sqrt{\frac{\gamma p}{\rho}}$.
$(i)$ From the relation $v = \sqrt{\frac{\gamma R T}{M}}$,it is clear that the velocity of sound is proportional to the square root of the absolute temperature,i.e.,$v \propto \sqrt{T}$.
(ii) From the relation $v = \sqrt{\frac{\gamma p}{\rho}}$,if the density $\rho$ is kept constant,the velocity of sound is proportional to the square root of the pressure $p$,i.e.,$v \propto \sqrt{p}$.
Comparing these with the given options,both $B$ and $C$ are physically correct statements based on the derived relations.

(B) The velocity of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass of the gas.
Note that the velocity of sound is independent of pressure for an ideal gas.
Given:
$T_1 = 273 + 20 = 293 \text{ K}$
$T_2 = 273 + 40 = 313 \text{ K}$
$v_1 = 344.2 \text{ m/s}$
Using the ratio:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
$v_2 = v_1 \sqrt{\frac{313}{293}}$
$v_2 = 344.2 \times \sqrt{1.06826}$
$v_2 \approx 344.2 \times 1.03356$
$v_2 \approx 355.75 \text{ m/s} \approx 356 \text{ m/s}$.

(D) The speed of a longitudinal wave in a metallic bar is given by $V = \sqrt{\frac{Y}{\rho}}$,where $Y$ is Young's modulus and $\rho$ is the density.
Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = \frac{1}{2} \frac{\Delta Y}{Y} - \frac{1}{2} \frac{\Delta \rho}{\rho}$.
Given $\frac{\Delta Y}{Y} \times 100 = 1\%$ and $\frac{\Delta \rho}{\rho} \times 100 = 0.5\%$.
Substituting these values,the percentage change in speed is $\frac{\Delta V}{V} \times 100 = \frac{1}{2}(1\%) - \frac{1}{2}(0.5\%) = 0.5\% - 0.25\% = 0.25\%$.
The change in speed $\Delta V = \frac{0.25}{100} \times 400 \ m/s = 1 \ m/s$.
The final speed $V_{final} = V + \Delta V = 400 + 1 = 401 \ m/s$.

(C) The velocity of sound in an ideal gas is given by $V = \sqrt{\frac{\gamma RT}{M}}$.
Since $\gamma$,$R$,and $M$ are constant,$V \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.
Given that the velocity doubles,if $V_1 = V_0$ at $T_1 = 0^{\circ} C = 273 \ K$,then $V_2 = 2V_0$ at $T_2 = (\alpha + 273) \ K$.
Using the ratio: $\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{V_0}{2V_0} = \sqrt{\frac{273}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{273}{T_2}$.
Therefore,$T_2 = 4 \times 273 = 1092 \ K$.
Since $T_2 = \alpha + 273$,we have $\alpha = 1092 - 273 = 819^{\circ} C$.