Wave Wavelength Calculation: Frequency & Speed

Understanding the Relationship Between Wave Speed, Frequency, and Wavelength

Understanding the relationship between wave speed, frequency, and wavelength is fundamental in physics, particularly when studying wave phenomena. Imagine dropping a pebble into a calm pond; ripples spread outwards, and these ripples are a perfect analogy for waves. These waves possess several key characteristics: their speed (how fast they travel across the water), their frequency (how many ripples pass a fixed point per second), and their wavelength (the distance between two consecutive crests or troughs of the ripple). In physics, these concepts are not just abstract ideas; they are measurable quantities that are intrinsically linked. The speed of a wave is essentially how quickly its energy propagates through a medium. Frequency, on the other hand, tells us about the rate at which the wave source is oscillating. Wavelength is the spatial period of the wave, describing its size. This intricate dance between speed, frequency, and wavelength is elegantly captured by a simple yet powerful equation: wave speed = frequency × wavelength. This equation, often represented as v = foldsymbol{\lambda}, is a cornerstone for solving many problems related to waves, from light waves and sound waves to electromagnetic radiation and even seismic waves. By understanding and applying this formula, we can predict and explain a vast array of physical phenomena. For instance, when light travels from one medium to another, like from air into glass, its speed changes, but its frequency typically remains constant. This change in speed directly affects its wavelength, causing the light to bend – a phenomenon known as refraction. Similarly, the pitch of a sound wave is determined by its frequency, while its loudness is related to its amplitude. The speed of sound, however, depends on the properties of the medium it travels through, such as temperature and density. Grasping this fundamental equation allows us to quantify these behaviors and delve deeper into the fascinating world of waves.

Applying the Wave Equation: A Practical Example

Let's dive into a practical example to solidify our understanding of the wave equation and how to use it. We're presented with a scenario involving a wave that has a specific frequency and speed. Our goal is to calculate its wavelength. The given information is: the frequency ($f$) of the wave is $6.1 imes 10^{14} Hz$, and its speed ($v$) is $3.00 imes 10^8 m/s$. The frequency is measured in Hertz (Hz), which is equivalent to cycles per second, and the speed is in meters per second (m/s). The wavelength, which we aim to find, will be in meters (m). The core principle we will employ is the wave equation: v = foldsymbol{\lambda}. To find the wavelength (oldsymbol{\lambda}), we need to rearrange this equation. By dividing both sides by the frequency ($f$), we get: oldsymbol{\lambda} = v / f. Now, we can substitute the given values into this rearranged equation. So, oldsymbol{\lambda} = (3.00 imes 10^8 m/s) / (6.1 imes 10^{14} Hz).

Performing the calculation: First, divide the numerical parts: 3.00 / 6.1 oldsymbol{\approx} 0.4918. Next, handle the powers of ten. When dividing powers of ten, we subtract the exponents: $10^8 / 10^{14} = 10^{(8-14)} = 10^{-6}$. Therefore, the wavelength is approximately $0.4918 imes 10^{-6} m$. It's common practice in scientific notation to have a single non-zero digit before the decimal point. To adjust $0.4918 imes 10^{-6} m$ into standard scientific notation, we move the decimal point one place to the right, which means we must decrease the exponent by one. This gives us $4.918 imes 10^{-7} m$. Rounding this to a reasonable number of significant figures based on the input values (two significant figures for frequency and three for speed), we get $4.9 imes 10^{-7} m$. This calculated wavelength represents the spatial extent of one complete cycle of the wave.

Interpreting the Wavelength Calculation and Matching with Options

Having performed the wavelength calculation, it's crucial to interpret the result and ensure it aligns with the provided multiple-choice options. Our calculated wavelength is approximately $4.9 imes 10^{-7} m$. This value signifies the distance between successive crests or troughs of the wave. The exponent $-7$ indicates that this is a very small distance, which is typical for waves like visible light, given the extremely high frequency of $6.1 imes 10^{14} Hz$. Frequencies of this magnitude are characteristic of visible light, where different wavelengths correspond to different colors. For example, violet light has a wavelength around $400 nm$ ($4 imes 10^{-7} m$), and red light has a wavelength around $700 nm$ ($7 imes 10^{-7} m$). Our calculated value falls right within this visible spectrum range, suggesting we are dealing with visible light. Now, let's compare our result with the given options:

A. $1.83 imes 10^{23} m$: This value is astronomically large and does not match our calculation or the typical scale of wave phenomena like light.

B. $9.91 imes 10^8 m$: This value is also very large, much larger than the size of the Earth, and does not correspond to our calculated wavelength.

C. $4.9 imes 10^{-7} m$: This option is remarkably close to our calculated value of $4.918 imes 10^{-7} m$. The slight difference is likely due to rounding in our calculation or the exact precision of the input values. This matches our result very well.

D. $2.77 imes 10^{-4} m$: While this is a small value, it is significantly larger than our calculated result and falls outside the typical range for visible light frequencies. It might correspond to other types of waves, but not the one described.

Based on our detailed calculation and comparison, option C, $4.9 imes 10^{-7} m$, is the correct answer. This exercise highlights the importance of accurately applying the wave equation and understanding the units and magnitudes involved in wave physics. The ability to correctly calculate and interpret wavelengths is essential for understanding phenomena ranging from the colors we see to the functioning of technologies like lasers and radio communication.

The Significance of Wavelength in the Electromagnetic Spectrum

Exploring the significance of wavelength in the electromagnetic spectrum provides further context and appreciation for the calculated value. The electromagnetic spectrum encompasses all types of electromagnetic radiation, ordered by frequency or wavelength. This spectrum includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Each of these forms of radiation travels at the speed of light in a vacuum (c oldsymbol{\approx} 3.00 imes 10^8 m/s), and their properties are dictated by their frequency and wavelength, linked by the fundamental equation c = foldsymbol{\lambda}. Our calculation resulted in a wavelength of $4.9 imes 10^{-7} m$. To better understand where this value fits, we can convert it to nanometers (nm), a common unit for visible light. Since $1 nm = 10^{-9} m$, our wavelength is $4.9 imes 10^{-7} m imes (1 nm / 10^{-9} m) = 490 nm$. This wavelength of $490 nm$ falls squarely within the visible light portion of the electromagnetic spectrum. Specifically, $490 nm$ light is perceived as blue-green or cyan color. This demonstrates how a specific frequency of light ($6.1 imes 10^{14} Hz$) corresponds to a particular color that our eyes can detect. The visible light spectrum ranges roughly from $400 nm$ (violet) to $700 nm$ (red). Therefore, the calculated wavelength of $490 nm$ is not just a number; it represents a tangible phenomenon – a specific color of light. This connection between a seemingly abstract physics equation and our everyday perception of color is a testament to the power and elegance of physics. Moreover, understanding these relationships is critical for fields like astronomy, where analyzing the light from distant stars and galaxies allows scientists to determine their composition, temperature, and motion. It's also vital in communications technology, where different parts of the electromagnetic spectrum are used for transmitting information, from Wi-Fi signals (microwaves) to cellular data (radio waves and microwaves). The precise wavelength and frequency of radiation determine its energy and how it interacts with matter, influencing everything from medical imaging (X-rays) to the therapeutic effects of sunlight (ultraviolet radiation).

Conclusion: Mastering Wave Calculations

In conclusion, mastering wave calculations like the one performed for wavelength is a crucial skill in physics. We started with a wave having a frequency of $6.1 imes 10^{14} Hz$ and a speed of $3.00 imes 10^8 m/s$. By rearranging the fundamental wave equation, v = foldsymbol{\lambda}, to solve for wavelength (oldsymbol{\lambda} = v/f), we were able to substitute the given values and compute the result. The calculation yielded a wavelength of approximately $4.9 imes 10^{-7} m$, which, when converted to nanometers, is $490 nm$. This value correctly corresponds to the blue-green region of the visible light spectrum, indicating that the wave in question is a form of visible light. Comparing our meticulously derived answer to the provided options, we confirmed that C. $4.9 imes 10^{-7} m$ is the accurate choice. This problem not only tests the ability to apply a formula but also encourages an understanding of the physical meaning of the quantities involved and their place within broader scientific contexts, such as the electromagnetic spectrum. Keep practicing these types of problems, and don't hesitate to consult reliable resources for further learning and clarification.

For more in-depth information on wave physics and the electromagnetic spectrum, you can explore resources like HyperPhysics or The Physics Classroom.