A 30.0-cm-long wire having a mass of 10.0 g is fixed at the two ends and is vibrated in its fundamental mode. A 50.0-cm-long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air = 340 m s^{−1}.

#### Solution

Given:

Mass of long wire* M* = 10 gm = 10 × 10^{−3}

Length of wire* l *= 30 cm = 0.3 m

Speed of sound in air *v *= 340 m s^{−1}

Mass per unit length \[\left( m \right)\] is

\[m = \frac{\text { Mass }}{\text { Unit length }}\]

\[ = 33 \times {10}^{- 3} \text { kg/m }\]

Let the tension in the string be *T*.

The fundamental frequency \[n_0\] for the closed pipe is

\[n_0 = \left( \frac{v}{4I} \right)\]

\[ = \frac{340}{2 \times 30 \times {10}^{- 2}}\]

\[ = 170 \text { Hz }\]

The fundamental frequency \[n_0\] is given by : \[n_0 = \frac{1}{2l}\sqrt{\frac{T}{m}}\]

On substituting the respective values in the above equation, we get :

\[170 = \frac{1}{2 \times 30 \times {10}^{- 2}} \times \sqrt{\frac{T}{33 \times {10}^{- 3}}}\]

\[ \Rightarrow T = 347 \text{ Newton }\]

Hence, the tension in the wire is 347 N.