Question

In case of a vibrating string, the frequency of the first overtone is equal to the frequency of the-
$\left( {\text{A}} \right)$ Fundamental
$\left( {\text{B}} \right)$ First harmonic
$\left( {\text{C}} \right)$ Second harmonic
$\left( {\text{D}} \right)$ Third harmonic

Answer 1

HINT: Here we have to find the correct length for wavelength.
Also compare the two formulae of frequency.
We will get the required answer.

Formulae used:
$\nu = n\lambda $, $\nu = \sqrt {\dfrac{T}{m}} $

Complete step by step solution:
For fundamental,
$l = \dfrac{\lambda }{2}$
$\Rightarrow \lambda = 2l$
Now, we know that
 $\nu = n\lambda $
Putting the value of $\lambda $ and we get,
$\Rightarrow \nu = n \times 2l$
Now we use the formula,
$\nu = \sqrt {\dfrac{T}{m}} $
Putting the value of $\nu $ and we get,
$n \times 2l = \sqrt {\dfrac{T}{m}} $
Take $n$ term as LHS and remaining as RHS we get,
$ \Rightarrow n = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $
For first overtone,
${\lambda _1} = l$
So we can write it as,
$\nu = {n_1}{\lambda _1} = {n_1}l$
But, $\nu = \sqrt {\dfrac{T}{m}} $
Putting the value of \[v\] and we get
${n_1}{\lambda _1} = \sqrt {\dfrac{T}{m}} $
Take ${n_1}$ term as LHS and remaining as RHS we get,
${n_1} = \dfrac{1}{{{\lambda _1}}}\sqrt {\dfrac{T}{m}} $
On multiplying numerator and denominator by 2, we get
${n_1} = 2\left[ {\dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} } \right]$
Since $n = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $
$\Rightarrow {n_1} = 2n$
Thus the correct option is option $\left( {\text{C}} \right)$.

Additional information:
Whenever a string is set into vibrations, the vibrations consist of a fundamental frequency and a few higher frequencies and overtones. Resonance causes a vibrating string to produce sound with constant frequency. An overtone is ant frequency higher than the fundamental frequency of a sound.

There are four properties that affect the vibrations in the string-
Length of the string- when the length of the string is varied, it vibrated with a different frequency.
Diameter of the string- if the strings are thicker, they vibrate slower and hence produce lower frequencies.
Tensions in the strings- tighter strings produce higher frequencies than lose strings.
Density of the string- dense material vibrates slowly than less dense ones and hence less dense strings produce high frequencies.

The formulae that we calculated above are widely used to tune numerous string instruments like guitar, sitar, veena etc. these are similar to the one with a hollow pipe hence, they are also used to tune wing pipe instruments like flute.

Note:Always remember to consider the length from node to node and not node to anti node. Once you find the relation between the wavelength and length, all you have to do is put it in the formula and calculate n.