Beats Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Define Beats.

Ans. When two sound waves of equal amplitude but slightly different frequencies superpose, the resultant wave is a single sinusoidal wave with a varying amplitude that goes from maximum to zero and back. The amplitude variation causes the variation of intensity called Beats.

Ques. What is Beat frequency?

Ans. When two sound waves of slightly different frequencies interfere, a resultant wave with varying amplitude is observed. The frequency with which the amplitude of the resultant wave rises and falls are called beat frequency.

Ques. Write the formula of the beat frequency.

Ans. The beat frequency is given by the difference between the frequencies of two interfering sound waves.

Let f₁ and f₂ be the frequencies of two sound waves, then the beat frequency is given by

f_beat = f₁ – f₂

Ques. The sound wave of frequency below 20 Hz is known as

1. Audible sound
2. Infrasonic
3. Supersonic
4. Ultrasonic

Ans. The correct option is b. Infrasonic.

Explanation: The sound wave is classified on the basis of frequencies. The sound wave of frequency below 20 Hz is known as Infrasonic, the sound wave of frequency above 20 kHz is known as ultrasonic and the sound wave whose frequency is between 20 Hz and 20 kHz is known as audible sound.

Ques. In music, the tone color or tone quality is defined using which property?

1. Frequency
2. Location
3. Timbre
4. Amplitude

Ans. The correct option is c. Timbre.

Explanation: Timbre is the perceived sound quality of a musical note, sound, or tone. It is also known as tone color or tone quality.

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Short Answers Questions [2 Mark Questions]

Ques. Why the speed of sound is greater in solids and liquids than in gases even though, they are denser than gases?

Ans. The velocity of the sound wave can be given by

\(v = \sqrt{\frac{modulus of elasticity}{density}}\)

Since solids and liquids have a large value of modulus of elasticity than gases. This factor more than compensates for their densities than gases. Therefore, the speed of

sound is greater in solids and liquids than in gases even though, they are denser than gases.

Ques. What is meant by wave interference?

Ans. The phenomenon when two or more waves have the same frequency or wavelength and have a constant initial phase difference between them, superpose in a medium is known as wave interference.

Ques. How wave interference is classified?

Ans. The interference of waves is classified into two types

• Constructive interference: It occurs when two waves having the same phase are superimposed on each other, then the amplitude of two waves gets added and the resultant wave of larger amplitude is obtained.
• Destructive interference: It occurs when two waves completely out of phase are superimposed on each other.

Ques. Define the threshold of hearing and the threshold of pain.

Ans. The minimum intensity of sound that a human ear can detect is called the threshold of hearing i.e. 10^-12 Wm^-2

The maximum intensity of sound that a human ear can detect is called the threshold of pain i.e. 1 Wm^-2

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Long Answers Questions [3 Mark Questions]

Ques. Calculate the velocity of sound in a gas in which two wavelengths 204 cm and 208 cm produce 20 beats in 6 seconds.

Ans. Given

• The wavelength of the first sound wave, λ₁ = 204 cm = 2.04 m
• The wavelength of the second sound wave, λ₂ = 208 cm = 2.08 m
• Beat frequency, f_beat = 20 beats in 6 seconds = 20/6 Hz

Let v be the velocity of the sound wave, then

• Frequency of the first sound wave, f₁ = v/λ₁ = v / 2.04
• Frequency of the second sound wave, f₂ = v/λ₂ = v / 2.08

We have, beat frequency, f_beat = f₁ - f₂

⇒ 20/6 = (v / 2.04) - (v / 2.08)

⇒ 3.34 = 0.49 v - 0.48 v

⇒ v = 3.34/ 0.01 = 334 m/s

Therefore, the velocity of sound in the given gas is 334 m/s.

Ques. Two identical wires are stretched by the same tension 101 N and each emits a node of frequency 202 Hz. If the tension in one wire is increased by 1 N, then what is the beat frequency?

Ans. The speed of the wave in a stretched wire is given by

\(v = \sqrt{\frac{T}{\mu}}\)

Where

• T is tension on the stretched wire
• μ is the density of the wire

Speed of the wave is also given by

v = fλ

Where

• f is the frequency of the wave
• λ is the wavelength

Therefore,

fλ = \(\sqrt{\frac{T}{\mu}}\)

Since both the wire is identical, therefore

f ∝ \(\sqrt{T}\)

⇒ \(\frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}}\) ⇒ f₂ = f₁ \(\sqrt{\frac{T_1}{T_2}}\)

Given the initial frequency, f₁ = 202 Hz and tension T₁ = 101 N

If tension on the wire is increased by 1 N, then final tension, T₂ = 101 + 1 = 102 N

New frequency of the wave on the wire is

f₂ = 202\(\sqrt{\frac{102}{101}}\) = 203

Hence, the frequency of the beat = f₂ - f₁ = 203 - 202 = 1 Hz

Ques. List the various properties of sound waves.

Ans. The properties of sound waves are

• Amplitude: The maximum displacement peak of the wave is known as amplitude.
• Frequency: The number of cycles per second a wave completed in one second is called its frequency.
• Wavelength: The distance between the two consecutive points in a wave that is in phase is called wavelength.
• Speed of the sound: The speed of the sound wave in air is 343 m/s in dry air, however, its speed will change in different mediums.

Ques. Two organ pipes closed at one end produce 5 beats per second in fundamental mode. If the ratio of their length is 10:11, then what are their frequencies?

Ans. The frequency (f) of a closed organ pipe in fundamental mode is given by

f = v/4L

Where

• v is the speed of the sound
• L is the length of the organ pipe

Since the speed of sound is constant in a particular medium, therefore

f ∝ 1/L

Let f₁ and f₂ be the frequencies of the two closed organ pipes and, L₁ and L₂ be their lengths. Then

f₁/f₂ = L₂/L₁

Given

• L₁/L₂ = 10/11 ⇒ L₂/L₁ = 11/10
• f₁ - f₂ = 5 beats per second ⇒ f₁ = 5 + f₂

Therefore

f₁/f₂ = 11/10 

⇒ (5 + f₂)/f₂ = 11/10

On solving the above equation, we get

f₂ = 50 Hz

Also, we have f₁ = 5 + f₂

⇒ f₁ = 5 + 50 = 55 Hz

Hence, the frequencies produced by two closed organ pipes are 55 Hz and 50 Hz.

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Very Long Answers Questions [5 Mark Questions]

Ques. A tuning fork produces 3 beats per second when sounded together with a fork of frequency 364 Hz. When the first fork is loaded with a little wax then the number of beats becomes two per second. What is the frequency of the first fork?

Ans. Let the frequency of the first tuning fork be f₁ and the frequency of the second tuning fork be f₂

Given

• f₂ = 364 Hz
• Beat produced, fbeat = 3 beats per second

Among the two frequencies f₁ and f₂, it is not given which one is higher, therefore two possible equations can be made for beat frequency

f₂ - f₁ = 3 or f₁ - f₂ = 3

⇒ 364 - f₁ = 3 or f₁ - 364 = 3

⇒ f₁ = 361 Hz or 367 Hz

Now when the first fork is loaded with a little wax number of beats becomes 2 per second. Therefore

f₂ - f₁ = 2 or f₁ - f₂ = 2

⇒ 364 - f₁ = 2 or f₁ - 364 = 2

⇒ f₁ = 362 Hz or 365 Hz

Now, we know when a tuning fork is loaded with wax its frequency decreases. Therefore if we consider the initial frequency of the tuning fork to be 361 Hz, then after loading of wax its frequency becomes 362 Hz or 365 Hz which is impossible.

Hence, the initial frequency of the tuning fork is 367 Hz, hence on loading the wax its frequency is reduced which may be 362 Hz or 365 Hz. Here both frequencies are less than 367 Hz.

Ques. 25 tuning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats per second. If the frequency of the first tuning fork is the octave of the last fork, then what is the frequency of the 21st fork?

Ans. Let the frequency of the first fork be f₁, the frequency of the second fork be f₂, and so on.

Given that any two successive forks produce 3 beats per second i.e.

⇒ f₁ - f₂ = 3

Therefore, the frequency of the second tuning fork

⇒ f₂ = f₁ - 3

The frequency of the third tuning fork will be

f₃ = f₂ - 3 = (f₁ - 3) - 3 = f₁ - 6

The frequency of the fourth tuning fork will be

f₄ = f₃ - 3 = (f₁ - 6) - 3 = f₁ - 9

In general, frequency of the nth tuning fork will be

f_n = f₁ - 3(n-1) …(i)

Therefore, the frequency of the 25th tuning fork will be

f₂₅ = f₁ - 3(25 - 1) = f₁ - 72 …(ii)

Given, the frequency of the first tuning fork is the octave of the last fork

⇒ f₁ = 2f₂₅ …(iii)

Therefore, equation (ii) becomes

f₂₅ = 2f₂₅ - 72

⇒ f₂₅ = 72 Hz

Also equation (iii) becomes

⇒ f₁ = 2 x 72 = 144 Hz

Using equation (i), the frequency of the 21st tuning fork will be

f₂₁ = f₁ - 3(21 - 1) = f₁ - 60

⇒ f₂₁ = 144 - 60 = 84 Hz

Hence, the frequency of the 21st fork will be 84 Hz.

Ques. A tuning fork A of an unknown frequency produces six beats per second with fork B of a known frequency 380 Hz. When fork A is waxed, the beat frequency decreases to 3 Hz. What is the frequency of fork A?

Ans. Let the frequency of the tuning fork A be f₁ and the frequency of the tuning fork B be f₂

Given

• Frequency of the tuning fork B, f₂ = 380 Hz
• Beat produced, f_beat = 6 beats per second

Among the two frequencies f₁ and f₂, it is not given which one is higher, therefore two possible equations can be made for beat frequency

f₂ - f₁ = 6 or f₁ - f₂ = 6

⇒ 380 - f₁ = 6 or f₁ - 380 = 6

⇒ f₁ = 374 Hz or 386 Hz

Now when the first fork A is waxed, the beat frequency becomes 3 Hz. Therefore

f₂ - f₁ = 3 or f₁ - f₂ = 3

⇒ 380 - f₁ = 3 or f₁ - 380 = 3

⇒ f₁ = 377 Hz or 383 Hz

Now, we know when a tuning fork is loaded with wax it frequency decreases. Therefore if we consider the initial frequency of the tuning fork A to be 374 Hz, then after loading of wax its frequency becomes 377 Hz or 383 Hz which is impossible because both frequencies are greater than 374 Hz.

Hence, the frequency of the tuning fork A is 386 Hz, hence on being waxed its frequency is reduced which may be 377 Hz or 383 Hz. Here both frequencies are less than 367 Hz.

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