# Problems Based on Beats due Sound Notes

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#### Example – 1:

• Two sound waves having wavelengths of 87 cm and 88.5 cm when superimposed produce 10 beats per second. Find the velocity of sound.
• Solution:
• Given: Wavelength of first wave = λ1 = 87 cm = 87 × 10-2 m,  Wavelength of second wave = λ2 = 88.5 cm = 88.5 × 10-2 m, No. of beats = 10 per second.
• To Find: Velocity of sound = v =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ2 >   λHence n1 >   n2

Ans:  Velocity of sound 513.3 m/s

#### Example – 2:

• Wavelengths of two sound waves in a gas are 2.0 m and 2.1m respectively. They produce 8 beats per second when sounded together. Calculate the velocity of sound in the gas and the frequencies of the two waves.
• Solution:
• Given: Wavelength of first wave = λ1 = 2.0 m,  Wavelength of second wave = λ2 = 2.1m, No. of beats = 8 per second.
• To Find: Velocity of sound = v =? Frequencies of notes =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ2 >   λHence n1 >   n2

Now n1 = v/λ1  =  336 /2.0 = 168 Hz

and n2 = v/λ2  =  336 /2.1 = 160 Hz

Ans:  Velocity of sound 336 m/s, The frequencies of note are 168 Hz and 160 Hz.

#### Example – 3:

• Two sound waves of lengths 1m and 1.01m produce 6 beats in two seconds when sounded together in the air. Find the velocity of sound in air.
• Solution:
• Given: Wavelength of first wave = λ1 = 1 m,  Wavelength of second wave = λ2 = 1.01m, No. of beats = 6 per two second = 3 per second.
• To Find: Velocity of sound = v =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ2 >   λHence n1 >   n2

Ans:  Velocity of sound 303 m/s

#### Example – 4:

• Two tuning forks of frequencies 320 Hz and 340 Hz produce sound waves of lengths differing by 6cm in a medium. Find the velocity of sound in the medium.
• Solution:
• Given: Frequency of first wave = n1 = 320 Hz,  Frequency of second wave = n2 = 340 Hz, Difference in wavelengths = 6 cm = 6 × 10-2 m
• To Find: Velocity of sound = v =?

We have v= n  λ, Hence  λ = v/ n

Now λ ∝ 1/ n   given n2 >   nHence λ1 >  λ2

Ans:  Velocity of sound 326.4 m/s

#### Example – 5:

• Wavelengths of two sound waves in air are 81/174m and 81/175 m. When these waves meet at a point simultaneously, they produce 4 beats per second. Calculate the velocity of sound in air.
• Solution:
• Given: Wavelength of first wave = λ1 = 81/174 m,  Wavelength of second wave = λ2 = 81/175 m, No. of beats = 4 per second.
• To Find: Velocity of sound = v =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ1 >   λHence n2 >   n1

Ans:  Velocity of sound 324 m/s

#### Example – 6:

• Wavelengths of two sound waves in air are 81/173m and 81/170 m. When these waves meet at a point simultaneously, they produce 10 beats per second. Calculate the velocity of sound in air.
• Solution:
• Given: Wavelength of first wave = λ1 = 81/173 m,  Wavelength of second wave = λ2 = 81/170m, No. of beats = 10 per second.
• To Find: Velocity of sound = v =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ2 >   λHence n1 >   n2

Ans:  Velocity of sound 270 m/s

#### Example – 7:

• Wavelengths of two notes in the air are 70/153 m and 70 /157 m. Each of these notes produces 8 beats per second with the third note of fixed frequency. What are the velocity of sound in air and the frequency of the third note?
• Solution:
• Given: Wavelength of first wave = λ1 = 70/153 m,  Wavelength of the second wave = λ2 = 70/157 m, No. of beats with the third note = 8 per second.
• To Find: Velocity of sound = v =? The frequency of the third note =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ1 >   λHence n2 >   n1

Let n be  the frequency of the third note, such that n2 > n >   n1

Given n2 –  n   = 8 ………. (1)

and  n  –  n1= 8  ………. (2)

Adding equations (1) and (2) we get

Now n1 = v/λ1  = 280 /(70/153) = 4 × 153 = 612 Hz

and n  = n1 + 8 = 512 + 8 = 520 Hz

Ans:  Velocity of sound 280 m/s and frequency of the third note is 620 Hz.

#### Example – 8:

• Wavelengths of two notes in air are 80/179 m and 80/177 m. Each note produces 4 beats per second with a third note of fixed frequency. Calculate the velocity of sound in air. Ans : (320 m/s )
• Solution:
• Given: Wavelength of first wave = λ1 =80/179 m,  Wavelength of the second wave = λ2 = 80/177 m, No. of beats with the third note = 4 per second.
• To Find: Velocity of sound = v =?

We have v= n  λ, Hence  n = v/ λ

Now n ∝ 1/ λ   given λ1 <   λHence n1 >   n2

Let n be  the frequency of the third note, such that n1 > n >   n2

Given n1 –  n   = 4 ………. (1)

and  n  –  n2= 4  ………. (2)

Adding equations (1) and (2) we get

Ans:  Velocity of sound 320 m/s

 Science > Wave Motion > You are Here