Question

Two stretched strings have lengths $l$ and $2l$ while tensions are $T$ and $4T$ respectively. If they are made of same material, the ratio of their frequencies is :

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

Answer 1

The correct option is C

$1:1$

Step 1: Given

Length of first string: $l_1=l$

Length of second string: $l_2=2l$

Tension in first string: $T_1=T$

Tension in second string: $T_2=4T$

Step 2: Formula Used

The frequency of a stretched string is $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$, where $l$ is length of string, $T$ is tension in string and $\mu$ is linear density of the string.

Step 3: Find the ratios of frequencies

Calculate the frequency in the first string using the formula

$$\begin{gathered} f=\frac{1}{2l}\sqrt{\frac{T}{\mu }} \\ \Rightarrow f_1=\frac{1}{2l_1}\sqrt{\frac{T_1}{\mu }} \\ \Rightarrow f_1=\frac{1}{2l}\sqrt{\frac{T}{\mu }} \end{gathered}$$

Calculate the frequency in the second string using the formula

$$\begin{gathered} f=\frac{1}{2l}\sqrt{\frac{T}{\mu }} \\ \Rightarrow f_2=\frac{1}{2l_2}\sqrt{\frac{T_2}{\mu }} \\ \Rightarrow f_2=\frac{1}{2\left(2l\right)}\sqrt{\frac{4T}{\mu }} \\ \Rightarrow f_2=\frac{1}{4l}\sqrt{\frac{4T}{\mu }} \end{gathered}$$

Calculate the ratio between the frequencies by dividing the first frequency by the second frequency

$$\begin{gathered} \frac{f_1}{f_2}=\frac{\frac{1}{2l}\sqrt{\frac{T}{\mu }}}{\frac{1}{4l}\sqrt{\frac{4T}{\mu }}} \\ =\frac{4l}{2l}\sqrt{\frac{T}{4T}} \\ =2\sqrt{\frac{1}{4}} \\ =1 \end{gathered}$$

Hence, the ratio of the frequencies is $f_1:f_2=1:1$.