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 :

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

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