Question

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 :

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

Answer 1

Answer: (B)

20.0 cm

Answer 2

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,$\frac{\lambda}{2} = \frac{20}{100} m$
= $20\, cm$