Let's break down this problem step by step to figure out how long it takes for the sound wave to travel the given distance. We are provided with the frequency and wavelength of the sound wave, and the total distance it needs to cover.

First, we need to find the speed of the sound wave. The speed (\(v\)) of a wave is related to its frequency (\(f\)) and wavelength (\(\lambda\)) by the formula:

\[v = f \times \lambda\]

The given frequency is 4 kHz, which is equal to 4000 Hz (since 1 kHz = 1000 Hz).

\[f = 4 \text{ kHz} = 4 \times 1000 \text{ Hz} = 4000 \text{ Hz}\]

The given wavelength is 30 cm. We need to convert this to meters (since 1 m = 100 cm).

\[\lambda = 30 \text{ cm} = \frac{30}{100} \text{ m} = 0.30 \text{ m}\]

Now, we can calculate the speed of the sound wave:

\[v = f \times \lambda = 4000 \text{ Hz} \times 0.30 \text{ m}\]

\[v = 1200 \text{ m/s}\]

So, the speed of the sound wave is 1200 meters per second.

Next, we need to find the time it takes to travel a distance of 2.4 km. The given distance is 2.4 km. We need to convert this to meters (since 1 km = 1000 m).

\[d = 2.4 \text{ km} = 2.4 \times 1000 \text{ m} = 2400 \text{ m}\]

The relationship between speed, distance, and time is:

\[v = \frac{d}{t}\]

We want to find the time (\(t\)), so we can rearrange the formula:

\[t = \frac{d}{v}\]

Now, substitute the values for distance and speed:

\[t = \frac{2400 \text{ m}}{1200 \text{ m/s}}\]

\[t = 2.0 \text{ s}\]

Therefore, it will take 2.0 seconds for the sound wave to travel 2.4 km.

Understanding Sound Wave Properties

Sound waves are mechanical waves that travel through a medium. Their properties like frequency, wavelength, speed, and time are interconnected.

• Frequency (\(f\)): The number of wave cycles that pass a point per second, measured in Hertz (Hz). Higher frequency means a higher pitch.
• Wavelength (\(\lambda\)): The distance between two consecutive identical points on a wave, such as crest to crest or trough to trough, measured in meters (m).
• Speed (\(v\)): How fast the wave travels through the medium, measured in meters per second (m/s). The speed of sound depends on the properties of the medium (like temperature and density).
• Time (\(t\)): The duration for which the wave travels a certain distance.
• Distance (\(d\)): The total length covered by the wave.

The fundamental relationship linking speed, frequency, and wavelength (\(v = f\lambda\)) is crucial for solving many wave problems. Similarly, the relationship between speed, distance, and time (\(v = d/t\)) is a basic principle of motion.

Step-by-Step Calculation for Sound Wave Travel Time

1. Identify the given values: frequency (\(f\)), wavelength (\(\lambda\)), and distance (\(d\)).
2. Ensure all units are consistent (e.g., meters, seconds, Hz). Convert units if necessary.
3. Calculate the speed of the wave (\(v\)) using the formula \(v = f \times \lambda\).
4. Use the calculated speed and the given distance to find the time (\(t\)) using the formula \(t = d / v\).

In this problem, we followed these steps precisely, converting kHz to Hz, cm to m, and km to m to ensure consistency before performing calculations.

Revision Table: Key Concepts in Waves

Additional Information on Sound Wave Speed

The speed of sound is primarily determined by the medium through which it travels. Sound travels faster in solids and liquids than in gases because the particles are closer together and can transmit vibrations more quickly. For example, the speed of sound in air at room temperature is about 343 m/s, in water it's about 1480 m/s, and in steel it's around 5960 m/s.

Temperature also affects the speed of sound in gases; it increases with increasing temperature. The frequency and wavelength of a sound wave can change when it enters a different medium, but the frequency usually remains constant (as it's determined by the source), causing the wavelength and speed to change proportionally.

In this specific problem, the speed was calculated from the frequency and wavelength, implying these were the properties of the wave in the medium it was traveling through for the 2.4 km distance.