Two identical guitar wires have a fundamental frequency of $$600$$ Hz, when kept under the same tension. The tension in a wire can be increased by tightening the tuning keys. During the tightening process, length of vibrating section of wire remains same. What percentage increase in the tension of one of the wires will lead to six beats per second when they are sounded together?

Question

Infinity Learn · Accepted Answer

$$v=12$$ℓTμ ∴     %$$T=2$$[% v]$$=26600$$×$$100=2$$%

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$v = \frac{1}{2 ℓ} \sqrt{\frac{T}{μ}} ∴ % T = 2 [% v] = 2 \frac{6}{600} \times 100 = 2 %$

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Travelling wave

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v=12ℓTμ ∴ %T=2[% v]=26600×100=2%

Ready to Test Your Skills?

free study material

$v = \frac{1}{2 ℓ} \sqrt{\frac{T}{μ}} ∴ % T = 2 [% v] = 2 \frac{6}{600} \times 100 = 2 %$