Physics - Wave on String Question with Solution | TestHub

The distance between consecutive antinodes in a stationary wave is $\frac{\lambda }{2}$, not $\lambda$.

In a pipe open at both ends, all harmonics (both even and odd) are present, with frequencies $f_n=n\frac{v}{2L}$ where $n=1,2,3,\dots$

In a pipe closed at one end, only odd harmonics are present, with frequencies $f_n=(2n-1)\frac{v}{4L}$ where $n=1,2,3,\dots$

In a stretched string fixed at both ends, the frequencies of vibration are $f_n=n\frac{v}{2L}$. The fundamental frequency is $f_1$ (first harmonic). The first overtone corresponds to the next higher frequency, which is $f_2=2f_1$. This is exactly the second harmonic. Thus, the frequency of the first overtone is the same as the second harmonic.