Question

Transverse waves are produced in a stretched wire. Both ends of the string are fixed. Let us compare between second overtone mode (in numerator) and fifth harmonic mode (in denominator). match the following two columns.

Answer 1

Second overtone means third harmonic and hence there will be three antinodes. In fifth harmonic there will be 5 antinodes.

A: Since $f\propto \frac{1}{\lambda }\propto \frac{n}{l}$$\Rightarrow \frac{f_2}{f_5}=\frac{3}{5}$

B. Number of nodes ratio is $\frac{2+2}{4+2}=\frac{4}{6}=\frac{2}{3}$

C: Number of antinodes ratio is $\frac{3}{5}$ since number of antinodes are proportional to the number of harmonics.

D: Wavelength is proportional to $\frac{1}{f}$, hence $\frac{{\lambda }_1}{{\lambda }_2}=\frac{5}{3}$.