$(A)$ The speed of sound determined from this experiment is $332 m s ^{-1}$
$(B)$ The end correction in this experiment is $0.9 cm$
$(C)$ The wavelength of the sound wave is $66.4 cm$
$(D)$ The resonance at $50.7 cm$ corresponds to the fundamental harmonic
$\frac{3 V}{4\left(L_1+e\right)}=f_0$ $. . . . . . (1)$
For $2^{\text {nd }}$ resonance
$\frac{5 V }{4\left( L _2+ e \right)}= f _0$ $. . . . . . (2)$
$\frac{3 V }{4\left( L _1+ e \right)}=\frac{5 V }{4\left( L _2+ e \right)}$
$3\left( L _2+ e \right)=5\left( L _1+ e \right)$
$2 e =3 L _2-5 L _1$
$e =\frac{3 \times 83.9-5 \times 50.7}{2}=\frac{251.7-253.5}{2}=\frac{-1.8}{2}=-0.9$
$\frac{3 V }{4\left( L _1+ e \right)}= f _0$
$V =\frac{500 \times 4 \times 49.8}{3 \times 100}=20 \times 16.6=332 m / s$
$\lambda=\frac{ V }{ f _0}=\frac{332}{500} \times 100=66.4 cm$

