Question

A stretched wire emits a fundamental note of $256\,{\text{Hz}}$. Keeping the stretching force constant and reducing the length of wire by $10\,{\text{cm}}$, the frequency becomes $320\,{\text{Hz}}$, the original length of the wire is:
A. $100\,{\text{cm}}$
B. $50\,{\text{cm}}$
C. $400\,{\text{cm}}$
D. $200\,{\text{cm}}$

Answer 1

Use the formula for frequency of the fundamental note:
$n = \dfrac{v}{{2l}}\sqrt {\dfrac{T}{\mu }}$, and the reduced length be $l - 10$. Put this length in the second case. Equate the two cases.

Complete step by step solution:
In this problem, you are asked to find the original length of the wire.
For this, we use the formula which gives frequency of the fundamental note:
$n = \dfrac{v}{{2l}}\sqrt {\dfrac{T}{\mu }}$ …… (1)
Where,
$n$ indicates frequency of the fundamental note.
$v$ indicates the speed of the sound wave.
$l$ indicates the length of the wire.
$T$ indicates the tension force in the wire.
$\mu$ indicates the mass per unit length (mass density of the wire).

For the first case:
Given,
Frequency is $256\,{\text{Hz}}$ .

Substituting this value in equation (1), we get:
$n = \dfrac{v}{{2l}}\sqrt {\dfrac{T}{\mu }}$
$256 = \dfrac{v}{{2l}}\sqrt {\dfrac{T}{\mu }}$ …… (2)

For the second case:
Given,
Frequency becomes $320\,{\text{Hz}}$ and,
The length of the wire becomes $10\,{\text{cm}}$ less than the original length.

Substituting this value in equation (1), we get:
$n = \dfrac{v}{{2l}}\sqrt {\dfrac{T}{\mu }}$
$320 = \dfrac{v}{{2\left( {l - 10} \right)}}\sqrt {\dfrac{T}{\mu }}$ …… (3)

Now, we divide equation (2) by equation (3):

$$\Rightarrow \dfrac{{256}}{{320}} = \dfrac{{\dfrac{v}{{2l}}\sqrt {\dfrac{T}{\mu }} }}{{\dfrac{v}{{2\left( {l - 10} \right)}}\sqrt {\dfrac{T}{\mu }} }} \\\ \Rightarrow \dfrac{{256}}{{320}} = \dfrac{{\dfrac{v}{{2l}}}}{{\dfrac{v}{{2\left( {l - 10} \right)}}}} \\\ \Rightarrow \dfrac{{256}}{{320}} = \dfrac{v}{{2l}} \times \dfrac{{2\left( {l - 10} \right)}}{v} \\\ \Rightarrow \dfrac{4}{5} = \dfrac{{l - 10}}{l} \\\$$

Now, we cross multiply the fractions involved above:

$$\dfrac{4}{5} = \dfrac{{l - 10}}{l} \\\ \Rightarrow 4l = 5l - 50 \\\ \Rightarrow l = 50\,{\text{cm}} \\\$$

Hence, the original length of the wire is $50\,{\text{cm}}$.

The correct option is (B).

Additional information:
Natural frequency, or basic frequency, also referred to simply as the fundamental frequency, is defined as the lowest frequency of a periodic waveform. An overtone is any frequency higher than a sound 's fundamental frequency.

Note: In this problem you are asked to find the original length of the wire. For this, the two frequencies are provided for two different cases. First one is the fundamental frequency. Always keep in mind that the material is the same for the two cases, hence the mass density will also remain the same, as it is the property of a material. Tension force is also the same, as provided by the question. Throughout the numerical, these two physical quantities will remain constant.