Question Video: Determining the Distance between the Centres of a Rarefaction and Compression | Nagwa Question Video: Determining the Distance between the Centres of a Rarefaction and Compression | Nagwa

# Question Video: Determining the Distance between the Centres of a Rarefaction and Compression Physics • Second Year of Secondary School

## Join Nagwa Classes

A sound source generates 90 oscillations each three seconds. Given that sound waves travel at 340 m/s in air, calculate the distance between the centers of a rarefaction and a successive compression.

02:56

### Video Transcript

A sound source generates 90 oscillations each three seconds. Given that sound waves travel at 340 meters per second in air, calculate the distance between the centers of a rarefaction and a successive compression.

To get started, let’s refresh our memories about the properties of sound waves. A sound wave is a longitudinal wave, which means that its medium — in this case, air — oscillates parallel to the direction of overall wave propagation. We can draw a diagram to help illustrate this. Sound waves travel through a medium like air by means of pressure differences that cause air molecules to oscillate back and forth along the same direction that the wave is traveling.

Rarefaction refers to the areas of the wave with low particle density, and compression refers to the areas of high particle density. We should recall that the wavelength of such a wave, represented by 𝜆, refers to the distance between either successive centers of compression or successive centers of rarefaction. But in this question, we’ve been asked to find the distance between the centers of a rarefaction and a successive compression, so that would be half of a wavelength, or 𝜆 divided by two.

Now, let’s look at the values we were given in the question. We’re told the speed of a sound wave as well as the number of oscillations this wave completes in three seconds. With this information, we can calculate the wave’s frequency. Frequency, 𝑓, gives the number of complete wave cycles per second, so we have that 𝑓 equals 90 cycles divided by three seconds, which is equal to 30 cycles per second, or 30 hertz. At this point, it’ll be helpful to recall the wave speed formula: 𝑣 equals 𝑓 times 𝜆, where 𝑣 is the speed of the wave, 𝑓 is its frequency, and 𝜆 is its wavelength.

To answer this question, we’ll need to solve for the wavelength, so let’s rearrange this equation to make 𝜆 the subject. To do this, we simply divide both sides by 𝑓 so that frequency cancels out of the numerator and denominator, leaving 𝜆 by itself. Thus, the expression can be written as 𝜆 equals 𝑣 divided by 𝑓. We also need to remember that this question is asking for half of a wavelength. We really want to find 𝜆 divided by two, so let’s divide both sides of this equation by two.

We can now substitute in the values for 𝑣 and 𝑓. But before we calculate, let’s take a moment to check out the units here. We have meters per second divided by hertz. Recall that hertz is equivalent to inverse seconds. So, units of per second will cancel out of the numerator and denominator, leaving only meters. This is a good sign, since wavelength is a distance measurement.

Finally, grabbing a calculator, we get a result of 5.666 and so on meters. Choosing to round this to two decimal places, we have our final answer. The distance between the centers of a rarefaction and a successive compression in this sound wave is 5.67 meters.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions