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Stationary waves

Question

A wire having a linear density of 5 g/m is stretched between two rigid supports with a tension of 450 N. It is observed that the wire resonates at a frequency of 420 Hz. The next highest frequency at which the same wire resonates is 490 Hz. Find the length of the wire. (in m)

Moderate
Solution

Let the frequency 420 Hz corresponds to pth harmonic. The formula for pth harmonic is given by

        np=p2lTμ

Hence, 420=p2lTμ          …(i)

For the next higher frequency p is (p + 1), hence

          490=p+12lTμ        …(ii)

From Eqs. (i) and (ii), we have 490420=p+1p

Solving we get, p = 6

Substituting the value p in Eq. (i), we get

420=62l4505×1031/2l=2.14 m



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Similar Questions

Two metallic strings A and B of different materials are connected in series forming a joint. The strings have similar cross-sectional area a = 1 mm2. The length of A is lA=0.3m and that B is lB=0.75m. One end of the combined string is tied with a support rigidly and the other end is loaded with a block of mass m = 25.2 kg passing over a frictionless pulley. Transverse waves are set up in the combined string using an external source of variable frequency, calculate the lowest frequency (in Hz) for which standing waves are observed such that the joint is a node.

The densities of A and B are 6.3 x 103 kg/m3 and 2.8 x 103 kg/m3 respectively.


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