A violin string $\mathbf{15}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$long and fixed at both ends oscillates in its $\mathit{n}\mathbf{=}\mathbf{1}\mathbf{}$mode. The speed of waves on the string is $\mathbf{250}\mathbf{\text{\hspace{0.17em}m/s}}\mathbf{}$, and the speed of sound in air is $\mathbf{348}\mathbf{\text{\hspace{0.17em}m/s}}$.

What are the (a) frequency and (b) wavelength of the emitted sound wave?

1. Speed of wave on the string,$\mathrm{v}=250\text{\hspace{0.17em}m/s}$
2. $\mathrm{n}=1$
   
3. Speed of sound in air,${\mathrm{v}}_{\mathrm{a}}=348\text{\hspace{0.17em}m/s}$
4. Length of the violin string,$\mathrm{l}=15\text{\hspace{0.17em}cmor}0.15\text{\hspace{0.17em}m}$

A periodic motion of a point down a straight line with acceleration always toward a fixed point in that line and a distance from that point proportional to that acceleration is known as a SHM. Here,$\mathbf{n}\mathbf{=}\mathbf{1}$ is for the fundamental frequency. So, the length of the string is half the wavelength of the wave on the string. Again, using the direct formula of frequency, we obtain the frequency of the emitted wave.

Formulae:

The wavelength of fundamental frequency (n=1), $\mathrm{\lambda }=2\mathrm{l}$ …(i)

The frequency of an oscillation,$\mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }}$ …(ii)

The wavelength of an oscillation in air, $\mathrm{\lambda }=\frac{{\mathrm{v}}_{\mathrm{a}}}{\mathrm{f}}$ …(iii)

Wavelength on string for the fundamental frequency can be given using the equation (i) as:

$\begin{array}{c}\mathrm{\lambda }=0.15\text{\hspace{0.17em}m}\left(2\right)\\ =0.3\text{m}\end{array}$

Frequency of an oscillation using equation (ii) can be written as:

$\begin{array}{c}\mathrm{f}=\frac{\frac{250\text{\hspace{0.17em}m}}{\text{s}}}{0.3\text{\hspace{0.17em}m}}\\ =833.3\text{Hz}\end{array}$

Frequency of vibration of the string will be the same for the frequency of sound.

Hence, the frequency of the emitted wave is $833.3\text{Hz}$

Wavelength of the emitted wave in air medium using equation (iii) can be written as:

$\begin{array}{c}\mathrm{\lambda }=\frac{\frac{348\text{\hspace{0.17em}m}}{\text{s}}}{833.3\text{\hspace{0.17em}Hz}}\\ =0.418\text{m}\end{array}$

Hence, the value of the wavelength of the emitted wave is $0.418\text{m}$