A Pipe open at one end has length 0.8 m. At the open end of the tube a string 0.5 m long is vibrating in its first overtone and resonates with fundamental frequency of pipe. If tension in the string i

Question

A Pipe open at one end has length 0.8 m. At the open end of the tube a string 0.5 m long is vibrating in its first overtone and resonates with fundamental frequency of pipe. If tension in the string is 50 N, the mass of string is (Neglect end correction) (Speed of sound = 320 m/s)

TestHub · Accepted Answer

The fundamental frequency of a pipe open at one end (closed pipe) is given by $f_{p} = \frac{v}{4 L _{p}}$ .

Substituting the given values, $f_{p} = \frac{320}{4 \times 0.8} = 100$ Hz.

The string is vibrating in its first overtone, which corresponds to the second harmonic. The frequency of the string is $f_{s} = \frac{2}{2 L _{s}} \frac{T}{μ} = \frac{1}{L _{s}} \frac{T}{μ}$ .

Since the string resonates with the pipe, $f_{s} = f_{p}$ .

$\frac{1}{0.5} \frac{50}{μ} = 100$

$2 \frac{50}{μ} = 100 \Rightarrow \frac{50}{μ} = 50$

Squaring both sides, we get $\frac{50}{μ} = 2500 \Rightarrow μ = \frac{50}{2500} = 0.02$ kg/m.

The mass of the string is $m = μ L_{s} = 0.02 \times 0.5 = 0.01$ kg.

Converting to grams, $m = 10$ g.

Answer: $10$ gram

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