Question

Two parts of a sonometer wire divided by a movable bridge differ in length by $0.2cm$ and produce $1beat/s$ when sounded together. The total length of the wire is $1m$, then the frequencies are

• A. $250.5Hz$ and $249.5Hz$
• B. $230.5Hz$ and $229.5Hz$
• C. $220.5Hz$ and $219.5Hz$
• D. $210.5Hz$ and $209.5Hz$

Answer 1

The correct option is A

$250.5Hz$ and $249.5Hz$

Step 1: Given data

Difference in length: $l=l_1-l_2=0.2cm--------\left(1\right)$

Total length of wire: $L=l_1+l_2=1m=100cm-----\left(2\right)$

Step 2: Formula Used

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

Step 3: Solution

$$\begin{gathered} l=l_1-l_2=0.2cm----------\left(1\right) \\ L=l_1+l_2=100cm----------\left(2\right) \\ \left(1\right)+\left(2\right)\Rightarrow \\ {l}_1=50.1cm \\ {l}_2=49.9cm \end{gathered}$$

Given that the number of beats is 1 per second.

So,

$$\begin{gathered} f_2-f_1=1 \\ {f}_2=f_1+1 \end{gathered}$$

Frequency,

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Since $T$ and $\mu$ are constant, $fl$ is also constant

$$\begin{gathered} \Rightarrow fl=const \\ \Rightarrow f_1{l}_1=f_2{l}_2 \\ \Rightarrow f_1\left(50.1\right)=\left(f_1+1\right)\left(49.9\right) \\ \Rightarrow f_1=249.5Hz \\ \Rightarrow f_2=f_1+1=249.5+1=250.5Hz \end{gathered}$$

Thus, the corresponding frequencies are $249.5Hz$ and $250.5Hz.$

Hence, option A is correct.