Question

A string of length $1\text{m}$ and mass $5\text{g}$ is fixed at both ends. The tension in the string is $8.0\text{N}$. The string is set into vibration using an external vibrator of frequency $100\text{Hz}$. The separation between successive nodes on the string is close to:

• A. $10.0\text{cm}$
• B. $33.3\text{cm}$
• C. $16.6\text{cm}$
• D. $20.0\text{cm}$

Answer 1

The correct option is D $20.0\text{cm}$

Given:
String length $=1\text{m}$
mass $=5\text{g}$
tension in the string $=8.0\text{N}$.
vibrator frequency $=100\text{Hz}$.

Velocity of wave on string
$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{8}{5}\times 1000}=40\text{m/s}$

Here, $T=$ tension and
$\mu =$ mass/length
Wavelength of wave
$\lambda =\frac{v}{n}=\frac{40}{100}\text{m}$

Separation between successive nodes,

$\frac{\lambda }{2}=\frac{40}{2\times 100}=\frac{20}{100}\text{m}=20\text{cm}$

Hence, option $\left(D\right)$ is correct.