Question

$4\text{m}$ long rope, having a mass of $40\text{g}$ is fixed at one end and is tied to a light string at the other end. Find the tension in the string if the difference between frequencies of consecutive overtones is $40\text{Hz}$.

• A. $1024\text{N}$
• B. $512\text{N}$
• C. $2048\text{N}$
• D. $1000\text{N}$

Answer 1

The correct option is A $1024\text{N}$

From formula for string fixed at one end,
$f_n=\frac{2n+1}{4L}\sqrt{\frac{T}{\mu }}$
and for $(n+1{)}^{th}$ overtone,
$f_{n+1}=\frac{2(n+1)+1}{4L}\sqrt{\frac{T}{\mu }}=\frac{2n+3}{4L}\sqrt{\frac{T}{\mu }}$
From question,
$f_{n+1}-f_n=40\text{Hz}$
$\Rightarrow \frac{2n+3}{4L}\sqrt{\frac{T}{\mu }}-\frac{2n+1}{4L}\sqrt{\frac{T}{\mu }}=40$
$\Rightarrow \frac{1}{4L}\sqrt{\frac{T}{\mu }}[2n+3-(2n+1\left)\right]=40$
$\Rightarrow \frac{1}{2L}\sqrt{\frac{T}{\mu }}=40\Rightarrow T=(40\times 2L{)}^2\times \mu$
From given data,
$\mu =\frac{40\times {10}^{-3}}{4}={10}^{-2}\text{kg/m}$
$L=4\text{m}$
$\Rightarrow T=(40\times 2\times 4{)}^2\times {10}^{-2}=1024\text{N}$