Question
A hollow pipe of length $$0.8\,m$$ is closed at one end. At its open end a $$0.5\,m$$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $$50\,N$$ and the speed of sound is $$320\,m{s^{ - 1}},$$ the mass of the string is
A.
5 grams
B.
10 grams
C.
20 grams
D.
40 grams
Answer :
10 grams
Solution :
Frequency of $${2^{nd}}$$ harmonic of string = Fundamental frequency produced in the pipe
$$\eqalign{ & \therefore \,\,2 \times \left[ {\frac{1}{{2{l_1}}}\sqrt {\frac{T}{\mu }} } \right] = \frac{v}{{4{l_2}}} \cr & \therefore \,\,\frac{1}{{0.5}}\sqrt {\frac{{50}}{\mu }} = \frac{{320}}{{4 \times 0.8}} \cr & \therefore \,\,\mu = 0.02\,kg\,{m^{ - 1}} \cr} $$
The mass of the string $$ = \mu {l_1}$$
$$\eqalign{ & = 0.02 \times 0.5\,kg \cr & = 10\,g \cr} $$
Frequency of $${2^{nd}}$$ harmonic of string = Fundamental frequency produced in the pipe
$$\eqalign{ & \therefore \,\,2 \times \left[ {\frac{1}{{2{l_1}}}\sqrt {\frac{T}{\mu }} } \right] = \frac{v}{{4{l_2}}} \cr & \therefore \,\,\frac{1}{{0.5}}\sqrt {\frac{{50}}{\mu }} = \frac{{320}}{{4 \times 0.8}} \cr & \therefore \,\,\mu = 0.02\,kg\,{m^{ - 1}} \cr} $$
The mass of the string $$ = \mu {l_1}$$
$$\eqalign{ & = 0.02 \times 0.5\,kg \cr & = 10\,g \cr} $$