Question

An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 m/s when the gas temperature is 22.0\(^\circ\)C. For a certain experiment, you need to have the same oscillator produce sound of wavelength 28.5 cm in this gas. What should the gas temperature be to achieve this wavelength?

Short answer

To produce a wavelength of 28.5 cm, the gas temperature should be approximately 81°C.

Step-by-step solution

Understand the Problem

We are given an oscillator vibrating at a frequency of 1250 Hz and a sound speed of 325 m/s at a temperature of 22.0°C. We want to find the new gas temperature such that the wavelength is 28.5 cm.

Recall the Wave Equation

The fundamental relationship between speed, frequency, and wavelength is given by the equation \( v = f \lambda \), where \( v \) is the speed of the wave, \( f \) is the frequency, and \( \lambda \) is the wavelength.

Convert and Set Known Values

First, convert the wavelength from centimeters to meters: 28.5 cm = 0.285 m. We know the current speed \( v = 325 \) m/s, and the frequency \( f = 1250 \) Hz. We want the speed at the new wavelength.

Calculate the Required Speed for Desired Wavelength

Using \( v = f \lambda \), calculate the speed needed for the new wavelength:\[v = 1250 \times 0.285 = 356.25 \text{ m/s}\]

Use the Speed of Sound Equation in Gases

The speed of sound in an ideal gas is given by \( v = \sqrt{\gamma \frac{R T}{M}} \), where \( \gamma \) is the adiabatic index (which is constant), \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.

Determine the Temperature Change

Since the speed equation implies \( v \propto \sqrt{T} \), we have:\[\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}\]Where \( v_1 = 325 \) m/s and \( v_2 = 356.25 \) m/s:\[\frac{325}{356.25} = \sqrt{\frac{295.15}{T_2}}\]Solving for \( T_2 \):\[\left(\frac{325}{356.25}\right)^2 \times 295.15 = T_2\]Calculate:\[T_2 \approx 354.1 \text{ K}\]

Convert to Celsius

Convert the Kelvin temperature back to Celsius: \( T_2 = 354.1 - 273.15 = 80.95 \). Therefore, the gas temperature should be approximately 81°C.

Key concepts

Wave Equation

The wave equation is a fundamental tool in physics to understand how waves propagate. It connects the speed of a wave, its frequency, and its wavelength through a simple equation: \( v = f \lambda \). Here,

• \( v \) stands for the speed of the wave.
• \( f \) represents the frequency of the wave, which is how often the wave oscillates.
• \( \lambda \) is the wavelength, the distance between successive crests or troughs of a wave.

This relationship helps us calculate one of these quantities if we know the other two. For instance, if we know the frequency and the speed of a sound wave, we can easily find the wavelength by rearranging the equation to \( \lambda = \frac{v}{f} \). This principle is widely used in various fields such as acoustics, optics, and electrical engineering.
The equation not only applies to sound waves but also to all types of waves, including light waves and water waves, making it an especially versatile equation in the study of wave phenomena.

Speed of Sound

The speed of sound is the speed at which sound waves travel through a medium. In the context of our exercise, we are interested in how sound propagates in gases. The speed of sound in a medium depends on the medium's properties, such as its density and elasticity.
For gases, the speed of sound is governed by the equation:\[ v = \sqrt{\gamma \frac{R T}{M}} \]where

• \( \gamma \) is the adiabatic index, a constant specific to the gas.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.
• \( M \) is the molar mass of the gas.

This equation highlights that the speed of sound in a gas is related to its temperature, which affects the kinetic energy of the gas molecules. Faster-moving molecules (as a result of higher temperatures) will transmit sound waves more quickly. Thus, when temperature increases, so does the speed of sound, implying that temperature and speed have a direct proportional relationship.

Gas Temperature Effect

Temperature plays a crucial role in determining the speed of sound, especially in gases. As discussed, the speed of sound equation reveals that the speed is proportional to the square root of the temperature. This means if the temperature of a gas changes, the speed of sound through the gas will also change, impacting how sound is perceived and transmitted.
To quantify this, the relationship between speed and temperature can be expressed as:\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]Here, we see that if the initial speed of sound \( v_1 \) and the final speed \( v_2 \) are known, along with the initial temperature \( T_1 \), one can solve for the final temperature \( T_2 \) in Kelvin.
This relationship is vital in many scientific and industrial processes, where controlling the temperature can precisely alter how fast sound waves travel through a medium. It's also crucial in environmental studies and meteorology, as temperature variations can affect sound propagation in the atmosphere.

Wavelength Calculation

Wavelength is a critical aspect of understanding sound waves, as it represents the physical length of one cycle of a wave. Calculating wavelength involves rearranging the wave equation to solve for \( \lambda \), giving us:\[ \lambda = \frac{v}{f} \]In our specific exercise, we need to adjust the conditions to obtain a particular wavelength of 28.5 cm (0.285 m). This requires us to calculate the necessary speed at the given frequency to achieve this wavelength. By substituting the known frequency and desired wavelength into the rearranged wave equation, we can determine the new speed needed:\[ v = f \cdot \lambda \]Wavelength calculation is fundamental in various applications such as designing musical instruments, engineering sonar equipment, and studying the acoustics of different environments. It helps us understand how changes in frequency or speed affect the wave's physical length, allowing precise control over sound transmission.