Question

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

• A. $33.3cm$
• B. $10.0cm$
• C. $16.6cm$
• D. $20.0cm$

Answer 1

The correct option is D $20.0cm$

Given, tension in the string, $T=8N$
Mass of the string $M=5g=5\times {10}^{-3}kg$

Length of the string, $l=1m$

Linear mass density of the string,
$\mu =\frac{M}{l}=5\times {10}^{-3}kg/m$

Velocity of wave on string

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{8}{5}\times 1000}=40m/s$
Given, frequency of the vibrator,
$f=100Hz$
Now, wavelength of wave $\lambda =\frac{v}{f}=\frac{40}{100}m$
Separation between successive nodes,

$\frac{\lambda }{2}=\frac{40}{200}m=20cm$
Final answer: (b)