Question

Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the length of the strings is

• A. $1:2$
• B. $1:3$
• C. 1:4
• D. 1:5

Answer 1

The correct option is B

$1:3$

Step 1: Given parameters

Two uniform strings A and B.

Let,

The frequency of A be ${\mathrm{n}}_1$.

The frequency of B be ${\mathrm{n}}_2$.

Step 2: To find

We have to find the ratio of the length of the strings.

Step 3: Calculation

$\therefore 2{\mathrm{n}}_1=3{\mathrm{n}}_2$

or,

$\frac{2}{2{\mathrm{l}}_1}\sqrt{\frac{\mathrm{T}}{{\mathrm{m}}_1}}=\frac{3}{2{\mathrm{l}}_2}\sqrt{\frac{\mathrm{T}}{{\mathrm{m}}_2}}$

Here,

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{mass}}{\mathrm{length}} \\ \mathrm{m}=\frac{\mathrm{cross}\mathrm{sectional}\mathrm{area}\times \mathrm{l}\times \mathrm{\rho }}{\mathrm{l}} \\ \mathrm{m}=\mathrm{cross}\mathrm{sectional}\mathrm{area}\times \mathrm{\rho } \end{gathered}$$

Here,

The cross-sectional area is “a”.

The density of the material is “$\mathrm{\rho }$”.

Now,

$$\begin{gathered} \frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{2}{3}\sqrt{\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}} \\ =\sqrt{\frac{{\mathrm{a}}_2\mathrm{\rho }}{{\mathrm{a}}_1\mathrm{\rho }}} \end{gathered}$$

Since the density is the same, then

$\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{2}{3}\sqrt{\frac{{{\mathrm{r}}_2}^2}{{{\mathrm{r}}_1}^2}}$

By substituting the values, we get

$$\begin{gathered} =\frac{2}{3}\sqrt{{\left(\frac{1}{2}\right)}^2} \\ =\frac{1}{3} \end{gathered}$$

Therefore, option B is the correct choice.