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Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.

## What are the Different Characteristics of Sound Waves and Solved Examples

Sound is a longitudinal wave which consists of compressions and rarefactions travelling through a medium. A sound wave can be described completely by five characteristics: Wavelength, Amplitude, Time-period, Frequency and Velocity (or Speed). All these characteristics of a sound wave are described below.

Consider the longitudinal sound waves ABCDE and EFGHI formed by the vibrations of a tuning fork (see Figure 13). The first sound wave starts from the normal density position A and after going through compression along ABC and rarefaction along CDE, it returns to the normal density position E.

Thus, the portion ABCDE represents one complete sound wave. The next sound wave starts from point E and goes up to point I (and so on). Thus, the above Figure shows two complete sound waves : from A to £, and from E to I. We will now describe the wavelength, amplitude, time-period, frequency and velocity (or speed) of the sound wave.

### 1. Wavelength

The minimum distance in which a sound wave repeats itself is called its wavelength. In most simple words, it is the length of one complete wave. The wavelength is denoted by the Greek letter lambda, λ. In a sound wave, the combined length of a compression and an adjacent rarefaction is called its wavelength.

For example, in Figure, the length of compression is AC and that of the adjacent rarefaction is CE, so the combined length AE is equal to wavelength. In a sound wave, the distance between the centres of two consecutive compressions (or two consecutive rarefactions) is also equal to its wavelength.

In Figure, the distance between the centres of two consecutive compressions is PR, so the distance PR represents wavelength of the sound wave. Similarly, the distance between the centres of two consecutive rarefactions is QS, so the distance QS also represents wavelength. Please note that the distance between the centres of a compression and an adjacent rarefaction is equal to half the wavelength \(\left(\frac{\lambda}{2}\right)\).

For example, the distance PQ in Figure is half wavelength \(\left(\frac{\lambda}{2}\right)\). The SI unit

for measuring wavelength is metre (m). Sometimes, however, centimetre unit is also used for expressing wavelength.

### 2. Amplitude

When a wave passes through a medium, the particles of the medium get displaced temporarily from their original undisturbed positions. The maximum displacement of the particles of the medium from their original undisturbed positions, when a wave passes through the medium, is called amplitude of the wave. The term amplitude is, in fact, used to describe the size of the wave.

In Figure, PB is the amplitude of the wave. The amplitude of a wave is usually denoted by the letter A. The SI unit of measurement of amplitude is metre (m) though sometimes it is also measured in centimetres. It should be noted that the amplitude of a wave is the same as the amplitude of the vibrating body producing the wave.

### 3. Time-Period

The time required to produce one complete wave (or cycle) is called time-period of the wave. Now, one complete wave is produced by one full vibration of the vibrating body. So, we can write another definition of time-period as follows : The time taken to complete one vibration is called time-period. Figure shows two complete waves (one wave from A. to E and another wave from E to I).

Suppose these two waves are produced in 1 second. Then the time required to produce one wave will be \(\frac{1}{2}\) second or 0.5 second. In other words, the time-period of this wave will be 0.5 second. The time-period of wave is denoted by the letter T. The unit of measurement of time

period is second (s).

### 4. Frequency

The term frequency tells us the rate at which the waves are produced by their source. The number of complete waves (or cycles) produced in one second is called frequency of the wave. Since one complete wave is produced by one full vibration of the vibrating body, so we can also say that:

The number of vibrations per second is called frequency. If 10 complete waves (or vibrations) are produced in one second, then the frequency of the waves will be 10 hertz (or 10 cycles per second). The frequency of a wave is fixed and does not change even when it passes through different substances.

The SI unit of frequency is hertz (which is written as Hz). A vibrating body emitting 1 wave per second is said to have a frequency of 1 hertz. In other words, 1 hertz is equal to 1 vibration per second. Sometimes, however, a bigger unit of frequency called kilohertz (kHz) is also used (1 kHz = 1000 Hz). The frequency of a wave is denoted by the letter f, though in some books, they use v (nu) to denote frequency.

The tuning forks are often marked with numbers like 256, 384 or 512, etc. These numbers signify the frequency of vibration of the tuning forks. For example, a tuning fork of frequency 256 means that its prongs will make 256 vibrations per second and emit 256 complete sound waves per second when hit on a hard surface. It should be clear by now that the frequency of a wave is the same as the frequency of the vibrating body which produces the wave.

We will now give the relation between time-period and frequency of a wave. The time required to produce one complete wave is called time-period of the wave. Suppose the time-period of a wave is T seconds.

Now, In T seconds, number of waves produced = 1

So, In 1 second, number of waves produced will be = \(\frac{1}{T}\)

But the number of waves produced in 1 second is called its frequency. This means that the frequency of the wave of time-period T will be \(\frac{1}{T}\). So, we can now say that the frequency of a wave is the reciprocal of its time-period. That is,

Frequency = \(\frac{1}{\text { Time period }}\)

or f =\(\frac{1}{T}\)

where f = frequency of the wave

and T = time-period of the wave

### 5. Velocity of Wave (or Speed of Wave)

The distance travelled by a wave in one second is called velocity of the wave (or speed of the wave). The velocity of a wave is represented by the letter v. The SI unit for measuring the velocity of a wave is metres per second (m/s or m s^{-1}).

### Relationship Between Velocity, Frequency and Wavelength of a Wave

We know that, Velocity = \(=\frac{\text { Distance travelled }}{\text { Time taken }}\)

Suppose a wave travels a distance lambda, λ, (which is its wavelength) in time T, then :

v = \(\frac{\lambda}{T}\)

Here T is the time taken by one wave. We know that \(\frac{1}{T}\) becomes the number of waves per second and this is known as frequency (f) of the wave. So, we can write f in place of \(\frac{1}{T}\) in the above relation. Thus,

v = f × λ

where v = velocity of the wave

f = frequency

and lambda, λ = wavelength

In other words : Velocity of a wave = Frequency × Wavelength

Thus, the velocity (or speed) of a wave in a medium is equal to the product of its frequency and wavelength. The formula v = f × λ, is called wave equation. It applies to all types of waves: transverse waves (like water waves), longitudinal waves (like sound waves) and electromagnetic waves (like light waves and radio waves).

The wave equation has three quantities in it, so if we know the values of any two quantities, then the value of third quantity can be calculated. We will use this formula to solve numerical problems.

**Example Problem 1.**

If 25 sound waves are produced per second, what is the frequency in hertz ?

**Solution:**

The frequency in hertz is equal to the number of waves produced per second. In this case, since 25 waves are being produced per second, so the frequency of the sound waves is 25 hertz (which is also written as 25 Hz).

**Example Problem 2.**

What is the frequency of a sound wave whose time-period is 0.05 s ?

**Solution:**

The relationship between the frequency and time-period of a wave is :

f = \(\frac{1}{T}\)

Here, Frequency, f = ? (To be calculated)

And, Time-period, T = 0.05 s

Putting this value in the above relation, we get:

f = \(\frac{1}{0.05}\)

f = \(\frac{100}{5}\)

f = 20 Hz

Thus, the frequency of the sound wave is 20 hertz.

**Example Problem 3.**

The wavelength of sound emitted by a source is 1.7 × 10^{-2} m. Calculate frequency of the sound, if its velocity is 343.4 m s^{-1}.

**Solution:**

The relationship between velocity, frequency and wavelength of a wave is given by the formula:

v = f × λ

Here, Velocity, v = 343.4 m s^{-1}.

Frequency, f = ? (To be calculated)

And, Wavelength, λ = 1.7 × 10^{-2} m

So, putting these values in the above formula, we get:

343.4 = f × 1.7 × 10^{-2}

f = \(\frac{343.4}{1.7 \times 10^{-2}}\)

f = \(\frac{3434 \times 10^2}{17}\)

f = 2.02 × 10^{4} Hz

Thus, the frequency of sound is 2.02 × 10^{4} hertz.

**Example Problem 4.**

Sound waves travel with a speed of about 330 m/s. What is the wavelength of sound whose frequency is 550 hertz ?

**Solution:**

Here, Speed of waves, v = 330 m/s

Frequency of waves, f = 550 Hz

And, Wavelength, λ = ? (To be calculated)

Ndw, v = f × λ

So, 330 = 550 × λ

λ = \(\frac{330}{550}\)

λ = 0.6 m

Thus, the wavelength of sound waves is 0.6 metre.

**Example Problem 5.**

A source is producing 1500 sound waves in 3 seconds. If the distance covered by a compression and an adjacent rarefaction be 68 cm, find : (a) frequency, (b) wavelength, and (c) velocity, of the sound wave.

**Solution:**

(a) Frequency. We know that frequency of a wave is the number of waves produced in 1 second.

Here, No. of waves produced in 3 seconds = 1500 .

So, No. of waves produced in 1 second = \(\frac{1500}{3}\) = 500

So, the frequency of this sound wave is 500 hertz.

(b) Wavelength. In a sound wave, the distance covered by a compression and an adjacent rarefaction is equal to its wavelength. This distance has been given to be 68 cm. So, the wavelength (X) of this sound wave is 68 cm.

(c) Velocity. The formula for calculating the velocity of a sound wave is :

v = f × λ

Here, Frequency, f = 500 Hz (Calculated above)

And, Wavelength, λ = 68 cm (Calculated above)

= \(\frac{68}{100}\) m

= 0.68 m

Putting these values of f and λ in the above formula, we get:

v = 500 × 0.68

= 340 m/s

Thus, the velocity (or speed) of the sound waves is 340 m/s.