A string of mass per unit length $\text{0.2} \mathrm{kg} {\mathrm{m}}^{-1}$ and length $\text{0.6} \mathrm{m}$ is fixed at both ends such that it has a tension of $80 \mathrm{N}$. If the string is vibrating in its second overtone mode, then the frequency of the string is

• A

  $\mathrm{40\; Hz}$
• Correct answer
  B

  $\mathrm{50\; Hz}$
• C

  $\mathrm{60\; Hz}$
• D

  $\mathrm{70\; Hz}$

Solution

$\text{v}=\sqrt{\frac{\text{T}}{\mu }}=\sqrt{\frac{\text{80}}{\text{0.2}}}=\text{20 m/s}$

$\frac{3\lambda }{2}=\text{0.6 m}$



$\mathrm{\therefore \; \; \; \; \lambda \; =\; 0.4\; m}$

$f=\frac{\text{v}}{\lambda }=\text{50 Hz}$

Hence, the correct answer is option: 'B'.