Q. Calculate the frequency of the second harmonic formed on a string of length $0.5\ m$ and mass $2 \times 10^{- 4}kg$ when stretched with a tension of $20\ N$
A) $274.4\ Hz$
B) $744.2\ Hz$
C) $44.72\ Hz$
D) $447.2\ Hz$
Explanation
(d) Mass per unit length $m = \frac{2 \times 10^{- 4}}{0.5}kg/m = 4 \times 10^{- 4}kg/m$
Frequency of $2^{nd}$ harmonic $n_{2} = 2n_{1}$
$= 2 \times \frac{1}{2l}\sqrt{\frac{T}{m}} = \frac{1}{0.5}\sqrt{\frac{20}{4 \times 10^{- 4}}} = 447.2Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.