Question

A $20\text{cm}$ long string, having a mass of $1.0\text{g}$, is fixed at both ends. The tension in the string is $0.5\text{N}$. The string is set into vibrations using an external oscillator of frequency $100\text{Hz}$. What is the seperation (in $\text{cm)}$ between the successive nodes on the string.

• A. $10\text{cm}$
• B. $5\text{cm}$
• C. $20\text{cm}$
• D. $15\text{cm}$

Answer 1

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

We know that,
Wave speed on a stretched string is given by,
$v=\sqrt{\frac{T}{\mu }}$
Where $\mu$ is mass per unit length of the string.

From the given data,
$v=\sqrt{\frac{0.5}{5\times {10}^{-3}}}=10\text{m/s}$

The wavelength of the wave established
$\lambda =\frac{v}{f}=\frac{10}{100}=0.1\text{m}\text{or}10\text{cm}$
The seperation between two successive nodes
$x=\frac{\lambda }{2}=\frac{10}{2}=5\text{cm}$

Thus, option (b) is the correct answer.