Question

The musical notes produced by an instrument are of frequencies $256Hz$ and $384Hz$ respectively. Compare their wavelengths. Given that speed of sound in air is $344m/s$.

Answer 1

Step 1 Given data:

1st frequency($f_1$) = $256Hz$

2nd frequency($f_2$) = $384Hz$

Speed of sound in air ($c$) = $344m/sec$

Wavelength of sound is $\lambda$

Step 2 Formula used:

$V=f\lambda$

where, $V$ is the speed of the sound, $\lambda$ the wavelength of sound and $f$ is the frequency of the sound.

Step 3 Finding wavelength:

$$\begin{gathered} 1stwavelength{\lambda }_1=\frac{V}{f_1}=344\times \frac{1}{256}=1.375m \\ 2ndwavelength{\lambda }_2=\frac{V}{f_2}=344\times \frac{1}{384}=0.8958m \end{gathered}$$

Hence, the wavelength of the first sound is more than the second sound.