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Home » A pipe open at one end has length . At the open end of the tube a string long is vibrating in its overtone and resonates with fundamental frequency of pipe. If tension in the string is , the mass of string is (speed of sound )A)25gB)15gC)20gD)10g
A pipe open at one end has length 0.8 \mathrm{~m}. At the open end of the tube a string 0.5 \mathrm{~m} long is vibrating in its 1^{\text {st }} overtone and resonates with fundamental frequency of pipe. If tension in the string is 50 \mathrm{~N}, the mass of string is (speed of sound =320 \mathrm{~m} / \mathrm{s})A)25gB)15gC)20gD)10g
Physics
A pipe open at one end has length 0.8 \mathrm{~m}. At the open end of the tube a string 0.5 \mathrm{~m} long is vibrating in its 1^{\text {st }} overtone and resonates with fundamental frequency of pipe. If tension in the string is 50 \mathrm{~N}, the mass of string is (speed of sound =320 \mathrm{~m} / \mathrm{s})A)25gB)15gC)20gD)10g

Correct option is 10g

Explanation:

Given, 2 nd harmonic of \mathrm{I}= Fundamental of II
\begin{array}{l} \therefore \quad 2\left(\frac{v_{1}}{2 l_{1}}\right)=\frac{v_{2}}{4 l_{2}} \Rightarrow \frac{T / \mu}{l_{1}}=\frac{v_{2}}{4 l_{2}} \\ \Rightarrow \quad \mu=\frac{16 T l_{2}^{2}}{v_{2}^{2} l_{1}^{2}}=\frac{16 \times 50 \times(0.8)^{2}}{(320)^{2} \times(0.5)^{2}} \\ \quad=0.02 \mathrm{~kg} / \mathrm{m} \end{array}
\therefore \quad m_{1}=\mu l_{1}=(0.02)(0.2)
=0.01 k g=10 g

i

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