Question

A dog produces a sound of frequency 30 kHz in air. (a) Taking the speed of sound in air to be 330 m st. calculate its wavelength. (b) What will be the distance between a rarefaction and its nearby compression ?​

Answer 1

Answer:

A dog produces a sound of frequency 30 kHz in the air if the speed of sound in air is 330 m/s, its wavelength is $11*10^{-3} m$ and the distance between rarefaction and nearby compression  is $5.5*10^{-3} m$

Explanation:

• In a sound wave, the high-density part is called compressions and rarefactions are the low-density parts of a wave

• The wavelength ( λ )of a sound wave is the distance between any two consecutive compressions or rarefactions
• For a sound wave moving with speed $v$ and frequency $f$ the wavelength is given by the formula  λ$=v/f$

• The distance between any compression and nearest rarefaction is half of the wavelength ( λ/2 ) as shown in the figure

From the question, we have

frequency of the sound produced by a dog $f=30kHz$

speed of that sound in air $v=330m/s$

substitute the values in the above equation to get the wavelength

⇒

a)  λ$=\frac{330}{30*10^{3} }= 11*10^{-3}m$

the distance between rarefaction and nearby compression is

b) λ/2$=\frac{11*10^{-3} }{2}=5.5*10^{-3}m$