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TS EAMCET · Physics · Waves and Sound

A wire of length \(0.4 \mathrm{~m}\) stretched at both ends vibrates 250 times per second. If the length of the wire is increased by \(0.1 \mathrm{~m}\) and the stretching force is reduced to \(1 / 4^{\text {th }}\) of its original value then the new frequency is

  1. A \(50 \mathrm{~Hz}\)
  2. B \(75 \mathrm{~Hz}\)
  3. C \(100 \mathrm{~Hz}\)
  4. D \(150 \mathrm{~Hz}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(100 \mathrm{~Hz}\)

Step-by-step Solution

Detailed explanation

We have \(\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}\) \(250=\frac{1}{2 \times 0.4} \sqrt{\frac{T}{\mu}}\)....(i) \(f^{\prime}=\frac{1}{2 \times 0.5} \sqrt{\frac{T}{4 \mu}}\)....(ii) Divide equation (ii) by (i), we get…
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