Question

A $20\text{cm}$ long string having a mass of $1\text{g}$ is fixed at both the ends. The tension in the string is $0.5\text{N}$. The string is set into vibrations using an external vibrator of frequency $100\text{Hz}$. The separation between the successive nodes on the string will be

• A. $4\text{cm}$
• B. $5\text{cm}$
• C. $6\text{cm}$
• D. $7\text{cm}$

Answer 1

The correct option is B $5\text{cm}$

Given:
Length of the string, $l=20\text{cm}=20\times {10}^{-2}\text{m}$
Mass of the string, $m=1\text{g}=1\times {10}^{-3}\text{kg}$
Tension in the string, $T=0.5\text{N}$
Frequency of the string, $f=100\text{Hz}$
We know that,
$v=\sqrt{\frac{T}{\mu }}$
$\Rightarrow v=\sqrt{\frac{0.5}{1\times {10}^{-3}/20\times {10}^{-2}}}=10\text{m/s}$
Also, $v=\lambda f$
$\Rightarrow 10=\lambda \times 100$
$\Rightarrow \lambda =0.1\text{m}$
We know that the distance between two successive nodes is $\frac{\lambda }{2}$.
So, $d=\frac{0.1}{2}=0.05\text{m}=5\text{cm}$

Why this question? Tips: In the case of a string fixed at both ends, all the harmonics of vibration are present. The string will vibrate in loops of length $\frac{\lambda }{2}$.