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Stationary waves

Question

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.

Moderate
Solution

Let p = number of loops formed in string A and q = number of loops formed in string B

Both p and q will be integer numbers, as the one end is fixed rigidly, the other end is loaded and the joint P is a node.

For string A, nA=p2lATμA=p2lATA

For string B, nB=q2lBTμB=q2lBTB

As both the strings are to be exerted by the same source of variable frequency, hence nA=nB.

          p2lATA=q2lBTB          pq=lAlBρAρB=0.30.756.3×1032.8×103           =2594=25×32=35nmin=32lATA=32×0.325.2×10106×6.3×103=1000 Hz



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