Question

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.

Answer 1

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 (m) is

m=MassUnit length=33×10−3kg/m

Let the tension in the string be T.

The fundamental frequency n0 for the closed pipe is

 n0=(v4I)=3402×30×10−2=170 Hz   

The fundamental frequency n0 is given by:

 n0=12lTm‾‾√   

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

170=12×30×10−2×T33×10−3‾‾‾‾‾‾‾√⇒ T=347 Newton

Hence, the tension in the wire is 347 N.

Answer 2

Answer:

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