KCET · Physics · Waves and Sound
A wire under tension vibrates with a fundamental frequency of \(600 \mathrm{~Hz}\). If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become
- A \(200 \mathrm{~Hz}\)
- B \(300 \mathrm{~Hz}\)
- C \(600 \mathrm{~Hz}\)
- D \(400 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(A) \(200 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Frequency of wire, \(f=\frac{1}{2 l r} \sqrt{\frac{T}{\pi \rho}}\)
So here,
\(\frac{f_{1}}{f_{2}} =\frac{l_{2}}{l_{1}} \times \frac{r_{2}}{r_{1}} \times \sqrt{\frac{T_{1}}{T_{2}}} \)
\(\frac{600}{f_{2}} =\frac{2}{1} \times \frac{1}{2} \times \sqrt{\frac{T}{T / 9}} \)
\(\frac{600}{f_{2}} =3 \)
\(f_{2} =\frac{600}{3} \)
\(f_{2} =200 \mathrm{~Hz}\)
So here,
\(\frac{f_{1}}{f_{2}} =\frac{l_{2}}{l_{1}} \times \frac{r_{2}}{r_{1}} \times \sqrt{\frac{T_{1}}{T_{2}}} \)
\(\frac{600}{f_{2}} =\frac{2}{1} \times \frac{1}{2} \times \sqrt{\frac{T}{T / 9}} \)
\(\frac{600}{f_{2}} =3 \)
\(f_{2} =\frac{600}{3} \)
\(f_{2} =200 \mathrm{~Hz}\)
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