Question

A stationary wave of frequency \(200 \mathrm{~Hz}\) are formed in air. If the velocity of the wave is \(360 \mathrm{~m} / \mathrm{s}\), the shortest distance between two antinodes is \(\ldots \ldots \ldots \ldots \mathrm{m}\) (A) \(1.8\) (B) \(3.6\) (C) \(0.9\) (D) \(0.45\)

Short answer

The shortest distance between two antinodes is \(0.9\) m.

Step-by-step solution

Recall and Write Down the Wave Formula

We know that the speed (velocity) of a wave is given by the product of its frequency (f) and wavelength (λ). The formula is v = λ × f where v is the velocity of the wave, λ is the wavelength, and f is the frequency. We are given v = 360 m/s and f = 200 Hz, and we need to find the wavelength (λ).

Solve for the Wavelength (λ)

Using the formula v = λ × f, we can rearrange it to solve for λ: λ = v / f Substitute the given values of v and f to find the wavelength: λ = (360 m/s) / (200 Hz) λ = 1.8 m

Determine the Relationship between Wavelength and Antinodes Distance

Now we have the wavelength of the wave, but we need to find the shortest distance between two antinodes. A stationary wave has one antinode at every half-wavelength. Therefore, the distance between two antinodes is equal to half of the wavelength.

Calculate the Shortest Distance between Two Antinodes

We know the distance between antinodes is equal to half of the wavelength. So, divide the wavelength by 2 to get the shortest distance between two antinodes: Shortest distance = λ / 2 Shortest distance = (1.8 m) / 2 Shortest distance = 0.9 m

Select the Correct Answer

Comparing our answer with the given options, the correct answer is (C) \(0.9\) m.

Key concepts

Wavelength

The wavelength ( \( \lambda \) ) of a wave is the distance over which the wave's shape repeats, typically measured from peak to peak or trough to trough. In the context of a stationary wave, such as the one described in the exercise, the wavelength signifies the distance between consecutive identical points on the wave considered as a total cycle.

In mathematical terms, the wavelength is determined by the formula \( v = \lambda \times f \) , where \( v \) stands for the velocity of the wave, \( \lambda \) is the wavelength, and \( f \) is the frequency.

When given the velocity (360 m/s) and frequency (200 Hz) of a wave, as in this exercise, you can find the wavelength by rearranging the formula: \( \lambda = \frac{v}{f} \). This shows that the wavelength is inversely proportional to the frequency for a constant velocity, providing crucial insight into how the physical properties can affect one another.

Antinodes

Antinodes in a stationary wave are the points where the wave achieves its maximum amplitude. Unlike nodes, which are points of no displacement, antinodes represent the most movement in a standing wave.

For a wave set into stationary motion, antinodes are significant because they illustrate where energy is concentrated and maximum. When dealing with problems involving stationary waves, understanding the location and distance between antinodes is essential for analyzing wave behavior accurately.

In this exercise, the distance between two consecutive antinodes is found by considering the structure of the wave. The distance between two antinodes reflects half of the wavelength, because, between one antinode and the next, a full crest and a full trough - which together make half perhaps - are fit. Understanding this concept helps in deeper comprehension of the wave's formation and pattern.

Wave Velocity

Wave velocity ( \( v \) ) is how fast the wave travels through a medium, and is crucial in defining how quickly the energy of the wave propagates. It can be affected by the properties of the medium it moves through. In many equations, wave velocity is a primary component, letting us determine other wave characteristics.

As outlined by the relation \( v = \lambda \times f \) , velocity is the product of the wavelength and the frequency. This means that if the wave's velocity is known (such as 360 m/s in our exercise), alongside frequency, calculating wavelength becomes straightforward, hence offering insight into the wave's behavior and the medium's characteristics.

Adjusting either the frequency or the medium's properties will adjust the velocity, demonstrating the interconnectedness of wave characteristics. This principle underscores much of wave dynamics and stationary wave formation.

Wave Frequency

Frequency ( \( f \) ) represents how many cycles of a wave pass a point in one second, measured in hertz (Hz). It defines the rate at which the wave oscillates and is a critical parameter in understanding wave mechanics.

In the given problem, the wave's frequency is 200 Hz, meaning each second, 200 cycle completions occur at any point. This frequency helps calculate other properties, particularly wavelength, when paired with velocity.

The relationship \( v = \lambda \times f \) demonstrates that wave frequency is inversely related to wavelength when velocity remains constant. This provides a basis for figuring out characteristic features of the wave.

Studying frequency not only aids in theoretical calculations but also in practical applications across fields like acoustics and signal processing, highlighting its ubiquity in understanding waves.