a) Discuss the effect of temperature and pressure on velocity of sound in a gas. [2] b) A train is approaching a cliff at 10 m/s. The driver sounds a whistle of frequency 600 Hz. What will be the…

a) Effect of Temperature and Pressure on Velocity of Sound in a Gas

The velocity of sound in a gas depends on several factors, including temperature and pressure. The key relationship is given by the formula for the speed of sound in an ideal gas:

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

where:

• $v$ is the speed of sound,
• $\gamma$ is the adiabatic index (ratio of specific heats, $C_p/C_v$),
• $R$ is the universal gas constant,
• $T$ is the absolute temperature (in Kelvin),
• $M$ is the molar mass of the gas.

Effect of Temperature:

• The speed of sound in a gas is directly proportional to the square root of the absolute temperature ($v\propto \sqrt{T}$).
• As temperature increases, the kinetic energy of gas molecules increases, leading to faster collisions and thus a higher speed of sound.
• For example, in air, the speed of sound increases by approximately 0.6 m/s for every 1°C rise in temperature.

Effect of Pressure:

• The speed of sound in a gas is independent of pressure (for a given temperature and gas composition).
• This is because pressure affects the density of the gas, but the ratio $\frac{\gamma P}{\rho }$ (where $\rho$ is density) remains constant for an ideal gas at constant temperature.
• However, if pressure changes while temperature remains constant, the density changes proportionally, leaving the speed of sound unchanged.

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b) Frequency of Echo Heard by the Driver

Given:

• Speed of the train ($v_s$) = 10 m/s (approaching the cliff),
• Frequency of the whistle ($f_0$) = 600 Hz,
• Speed of sound in air ($v$) = 340 m/s.

Step 1: Frequency of the echo as heard by the cliff (stationary observer) Since the train is moving towards the cliff, the frequency received by the cliff ($f_1$) is higher due to the Doppler effect:

$$f_1=f_0\left(\frac{v}{v-v_s}\right)$$

$$f_1=600\left(\frac{340}{340-10}\right)$$

$$f_1=600\left(\frac{340}{330}\right)$$

$$f_1=600\times 1.0303$$

$$f_1=618.18\text{Hz}$$

Step 2: Frequency of the echo as heard by the driver The echo reflects off the cliff and travels back towards the moving train. Now, the train is moving towards the source (the echo), so the observed frequency ($f_2$) is again increased:

$$f_2=f_1\left(\frac{v+v_s}{v}\right)$$

$$f_2=618.18\left(\frac{340+10}{340}\right)$$

$$f_2=618.18\left(\frac{350}{340}\right)$$

$$f_2=618.18\times 1.0294$$

$$f_2=636.36\text{Hz}$$

Final Answer: The frequency of the echo as heard by the driver is 636.36 Hz.