MHT CET · Physics · Waves and Sound
The frequency of a stretched uniform wire of length \(L\) under tension is in resonance with the fundamental frequency of a closed pipe of same length. If the tension in the wire is increased by 8 N , it is in resonance with the first overtone of the same closed pipe. The initial tension in the wire is
- A 4 N
- B \(\frac{1}{2} \mathrm{~N}\)
- C 2 N
- D 1 N
Answer & Solution
Correct Answer
(D) 1 N
Step-by-step Solution
Detailed explanation
\(f_{w1} = f_{p,fund} \implies \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} = \frac{v}{4L} \implies \sqrt{T_1} = \frac{v\sqrt{\mu}}{2}\) \(f_{w2} = f_{p,1st\_overtone} \implies \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} = 3 \frac{v}{4L} \implies \sqrt{T_2} = \frac{3v\sqrt{\mu}}{2}\)
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