Understanding the Speed of Sound: How Frequency and Wavelength Play a Role

What is the speed of sound of a voice that produces a sound with a frequency of 110 Hz and a wavelength of 3.14 m?

• A. 300 m/sec
• B. 345.4 m/sec
• C. 480 m/sec
• D. 235 m/sec

Answer: (B)

The speed of sound can be calculated using the formula that relates speed (v), frequency (f), and wavelength (λ): \[ v = f \times \lambda \] In this case, the frequency of the sound is given as 110 Hz and the wavelength is 3.14 m. Plugging these values into the formula: \[ v = 110 \, \text{Hz} \times 3.14 \, \text{m} \] Calculating this gives: \[ v = 345.4 \, \text{m/sec} \] This tells us that the correct speed of sound in this scenario is 345.4 m/sec. The results show that the speed of sound for a voice producing a frequency of 110 Hz with a wavelength of 3.14 m falls well within the range of typical sound speeds observed in air under normal conditions, which is generally around 343 m/sec at room temperature. In contrast, other options present values that do not match the calculated result based on the frequency and wavelength provided. These different speeds could represent other conditions, mediums, or errors in calculation, but they do not accurately represent the speed derived from the relationships of frequency and wavelength specified in the problem.

When it comes to understanding sound, it’s not all about how loud or soft something is. It's also about how fast those waves travel, which brings us to an exciting topic: the speed of sound! You might be asking yourself: how do we figure that out? Well, let's break it down step by step, focusing on a particular example that ties into the PSC1121 curriculum at UCF.

So, imagine you hear a voice producing a sound with a frequency of 110 Hz and a wavelength of 3.14 m. What’s the speed of sound in this case? Is it 300 m/sec? Maybe 480 m/sec? Or is it something else entirely? The correct answer is actually 345.4 m/sec. How do we get there? Great question!

We can use the relationship laid out in a nifty formula:

[ v = f \times \lambda ]

Where:

• ( v ) is the speed of sound,

• ( f ) is the frequency (that’s the 110 Hz we’ve got), and

• ( \lambda ) is the wavelength (3.14 m in this example).

Here’s the thing: when we plug in these numbers:

[ v = 110 , \text{Hz} \times 3.14 , \text{m} ]

You get:

[ v = 345.4 , \text{m/sec} ]

This is a fantastic demonstration of how frequency and wavelength interact, creating pathways for sound to travel through our air. In typical conditions—say at room temperature—sound travels at about 343 m/sec. So, our calculated speed fits perfectly within what you'd expect!

You might wonder, how do the other options stack up? Well, the values of 300 m/sec, 480 m/sec, and 235 m/sec fall outside the expected range for sound traveling through air, and these could imply either different media (like water or steel) or simply mistakes in calculation.

Understanding this relationship is critical, not just for exam purposes but for grasping how sound waves function in our world. Think about everyday experiences: when you're at a concert, the sound doesn’t just pop into existence; it travels through the air to reach your ears. Sound is more than vibrations; it’s a symphony of physics at work!

So next time you're studying for that exam, remember, you’re not just memorizing numbers and equations — you are learning about the fundamental principles that govern how we experience sound in our lives. Keep these concepts close to your heart (and your study notes), and you’ll have a much easier time tackling the questions that come your way!