Question

If equation of sound wave is $\mathrm{y}=0.0015\sin \left(62.4\mathrm{x}+316\mathrm{t}\right)$, then its wavelength will be ______.

Answer 1

Step 1. Given data

The equation of sound wave is, $\mathrm{y}=0.0015\sin \left(62.4\mathrm{x}+316\mathrm{t}\right)$

We have to find the wavelength.

Step 2. Formula to be used

The standard form of the sound wave is,

$\mathrm{y}=\mathrm{Asin}\left(\mathrm{kx}+\mathrm{\omega t}\right)$

$\mathrm{y}=\mathrm{Asin}2\mathrm{\pi }\left[\frac{\mathrm{x}}{\mathrm{\lambda }}+\frac{\mathrm{t}}{\mathrm{T}}\right]$

Here, $\mathrm{A}$ is amplitude, $\mathrm{\omega }$ is the angular frequency, $\mathrm{k}$ is the angular wavenumber, $\mathrm{t}$ is time and $\mathrm{x}$ is position.

Step 3. Find the wavelength of the sound wave.

Let,

$\mathrm{y}=0.0015\sin \left(62.4\mathrm{x}+316\mathrm{t}\right)$ ------ $\left(1\right)$

Now, the standard form of wave is,

$\mathrm{y}=\mathrm{Asin}\left(\mathrm{kx}+\mathrm{\omega t}\right)$ ------- $\left(2\right)$

Compare the above equation $\left(1\right)$ with the standard equation $\left(2\right)$.

Here,

$\mathrm{A}=0.0015$

$\mathrm{\omega }=316$

$\mathrm{k}=62.4$

We know that,

$\mathrm{k}=\frac{2\mathrm{\pi }}{\lambda }$

$62.4=\frac{2\times 3.14}{\mathrm{\lambda }}$

$\mathrm{\lambda }=\frac{2\times 3.14}{62.4}$

$\mathrm{\lambda }=\frac{157}{1560}$

$\mathrm{\lambda }=0.10$

Hence, if the equation of sound wave is $\mathrm{y}=0.0015\sin \left(62.4\mathrm{x}+316\mathrm{t}\right)$, then its wavelength will be $0.1$ unit.