# Question #2148b

Feb 24, 2017 Given that the velocity of stationary wave produced in a stretched string depending on its tension (T)and its mass per unit length (m) is $\left(V = 270.9 \text{m/s}\right)$

As shown in above figure the wave lengths ${\lambda}_{i}$ of first three harmonics are related with the length $L$ of the stretched string as follows.

For first i th harmonic ${\lambda}_{i} = \frac{2 L}{p} _ i$,

where ${p}_{i}$ is the number of loops produced in the given length L of the stretched string

For 1st harmonic wave length ${\lambda}_{1} = 2 L$

For 2nd harmonic wave length ${\lambda}_{2} = \frac{2 L}{2}$

For 3rdharmonic wave length ${\lambda}_{2} = \frac{2 L}{3}$

So

For 1st harmonic, frequency

${n}_{1} = \frac{V}{\lambda} _ 1 = \frac{V}{2 L} = \frac{270.9 \times {10}^{2} c m {s}^{-} 1}{32.4 c m} = 836.1 H z$

For 2nd harmonic, frequency

${n}_{2} = \frac{V}{\lambda} _ 2 = \frac{2 V}{2 L} = 2 {n}_{1} = 1672.2 H z$

For 3rd harmonic, frequency

${n}_{3} = \frac{V}{\lambda} _ 2 = \frac{3 V}{2 L} = 3 {n}_{1} = 2508.3 H z$