Physics - Sound Wave Question with Solution | TestHub

Let$n_1$harmonic is corresponding to$50.7 cm \mathrm{a}\mathrm{n}\mathrm{d} n_2$harmonic is corresponding$83.9 cm$. since both one consecutive harmonics.
$\therefore$There difference$=\frac{\lambda }{2}$
$\therefore \frac{\lambda }{2}=\left(83.9-50.7\right)cm$
$\frac{\lambda }{2}=33.2 cm$
$\lambda =66.4 cm$
$\therefore \frac{\lambda }{4}=16.6cm$
length corresponding to fundamental mode must be close to$\frac{\lambda }{4} \mathrm{a}\mathrm{n}\mathrm{d} 50.7 cm$must be an odd multiple of this length$16.6 \times 3 = 49.8 cm$. therefore$50.7 is 3^{rd}$harmonic
If end correction is e, then
$e+50.7=\frac{3\lambda }{4}$
$e=49.8-50.7=0.9cm$
Speed of sound,$c=f\lambda$
$\therefore v=500\times 66.4cm/sec=332.000m/sec$