Question

A $1\text{m}$ long horizontal rope, having a mass of $40\text{g}$, is fixed at one end and is tied to a light string at the other end. The tension in the rope is $400\text{N}$. What will be the wavelengths (in $\text{metres}$) in the first and second overtone ?

• A. $\frac{3}{4},\frac{3}{4}$
• B. $\frac{4}{3},\frac{4}{5}$
• C. $\frac{5}{4},\frac{5}{3}$
• D. $\frac{4}{5},\frac{4}{3}$

Answer 1

The correct option is B $\frac{4}{3},\frac{4}{5}$

Given,
Length of rope, $L=1\text{m}$
Mass, $M=40\text{g}\text{or}0.04\text{kg}$
Mass per unit length, $\mu =\frac{M}{L}=0.04\text{kg/m}$
Tension in the rope, $T=400\text{N}$

It behaves like a string fixed at one end.
$\therefore$ wave speed on the stretched string, $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{400}{0.04}}=100\text{m/sec}$

Fundamental frequency, $f_0$ on a string fixed at one end is given by
$f_0=\frac{v}{4L}$
$\Rightarrow f_0=\frac{100}{4}=25\text{Hz}$

Frequency of the first overtone on the string fixed at one end
$f_1=\frac{3v}{4L}$
$\Rightarrow$ Wavelength of first overtone ${\lambda }_1=\frac{v}{f_1}=\frac{4L}{3}$
From the given data,
${\lambda }_1=\frac{4\times 1}{3}=\frac{4}{3}\text{m}$

Frequency of the second overtone, $f_2$ on the string,
$f_2=\frac{5v}{4L}$
$\Rightarrow$ Wavelength of second overtone ${\lambda }_2=\frac{v}{f_2}=\frac{4L}{5}$
From the given data,
${\lambda }_2=\frac{4\times 1}{5}=\frac{4}{5}\text{m}$

Thus, option (b) is the correct answer.