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

JEE MainJEE Main 2019Waves

Solution:

Velocity of wave on string
$V = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{8}{5} \times1000} = 40m/s $
Now, wavelength of wave $ \lambda = \frac{v}{n} = \frac{40}{100} m $
Separation b/w successive nodes,$ \frac{\lambda}{2} = \frac{20}{100} m $
= $20\, cm$