Physics - Wave on String Question with Solution | TestHub

The fundamental frequency of a closed organ pipe of length $L_p$ is given by $f_p=\frac{v}{4L_p}$.

Substituting the given values, $v=320$ m/s and $L_p=0.8$ m:

$f_p=\frac{320}{4\times 0.8}=\frac{320}{3.2}=100$ Hz

The frequency of the fourth harmonic of a string of length $L_s$ fixed at both ends is given by $f_s=\frac{4}{2L_s}\sqrt{\frac{T}{\mu }}$, where $T$ is the tension and $\mu$ is the linear mass density.

Given that the string resonates with the fundamental frequency of the pipe, $f_s=f_p=100$ Hz.

Substituting $L_s=0.5$ m and $T=50$ N:

$100=\frac{4}{2\times 0.5}\sqrt{\frac{50}{\mu }}$

$100=4\sqrt{\frac{50}{\mu }}$

$25=\sqrt{\frac{50}{\mu }}$

Squaring both sides, we get:

$625=\frac{50}{\mu }\Rightarrow \mu =\frac{50}{625}=0.08$ kg/m

The mass of the string $m$ is the product of its linear mass density and length:

$m=\mu L_s=0.08\times 0.5=0.04$ kg

Converting to grams, $m=40$ g.

Answer: $40$ g