Harmonics differ due to speed ratio on equal-length strings

Steps:

• Frequency $f$ is identical on both strings.
• Wavelength $\lambda_{PM} = v_{PM}/f = 2 v_{MQ}/f = 2 \lambda_{MQ}$.
• For fixed-end stationary waves, length $L = n \lambda / 2$, so $n_{PM} \lambda_{PM} = n_{MQ} \lambda_{MQ}$.
• Thus $n_{MQ} = 2 n_{PM}$; lowest mode is $n_{PM}=1$, (two half-wavelengths on MQ).

Why A is correct:

• Shows second harmonic: nodes at M/Q and midpoint, two antinodes, matching $L = \lambda_{MQ}$ for fixed ends.

Why the others are wrong:

• B: Fundamental ($n=1$), one antinode; requires $\lambda_{MQ} = 2L$, but speed ratio demands shorter $\lambda_{MQ}$.

Final answer: A