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Science > Physics > Wave Motion > You are Here |

**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 = λ_{1}= 87 cm = 87 × 10^{-2}m, Wavelength of second wave = λ_{2}= 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 λ_{2} > λ_{1 }Hence n_{1} > n_{2}

**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 = λ_{1}= 2.0 m, Wavelength of second wave = λ_{2}= 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 λ_{2} > λ_{1 }Hence n_{1} > n_{2}

Now n_{1} = v/λ_{1} = 336 /2.0 = 168 Hz

and n_{2} = v/λ_{2} = 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 = λ_{1}= 1 m, Wavelength of second wave = λ_{2}= 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 λ_{2} > λ_{1 }Hence n_{1} > n_{2}

**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 = n_{1}= 320 Hz, Frequency of second wave = n_{2}= 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 n_{2} > n_{1 }Hence λ_{1} > λ_{2}

**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 = λ_{1}= 81/174 m, Wavelength of second wave = λ_{2}= 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 λ_{1} > λ_{2 }Hence n_{2} > n_{1}

**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 = λ_{1}= 81/173 m, Wavelength of second wave = λ_{2}= 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 λ_{2} > λ_{1 }Hence n_{1} > n_{2}

**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 = λ_{1}= 70/153 m, Wavelength of the second wave = λ_{2}= 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 λ_{1} > λ_{2 }Hence n_{2} > n_{1}

Let n be the frequency of the third note, such that n_{2} > n > n_{1}

Given n_{2} – n = 8 ………. (1)

and n – n_{1}= 8 ………. (2)

Adding equations (1) and (2) we get

Now n_{1} = v/λ_{1} = 280 /(70/153) = 4 × 153 = 612 Hz

and n = n_{1} + 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 = λ_{1}=80/179 m, Wavelength of the second wave = λ_{2}= 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 λ_{1} < λ_{2 }Hence n_{1} > n_{2}

Let n be the frequency of the third note, such that n_{1} > n > n_{2}

Given n_{1} – n = 4 ………. (1)

and n – n_{2}= 4 ………. (2)

Adding equations (1) and (2) we get

**Ans: ** Velocity of sound 320 m/s

Science > Physics > Wave Motion > You are Here |

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