The total length of a sonometer wire between fixed ends is 110 cm… | JEE Main 2014 (19 Apr, Shift 1)

Q102JEE Main 2014 · 19 APR · SHIFT 1Medium

Stationary Waves & Vibrating Strings

The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6: 3: 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?

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Answer: (b)

Lengths: $60, 30, 20\text{ cm}$ . Wave speed $v=\sqrt{T/\mu}=\sqrt{400/0.01}=200\text{ m/s}$ .
Fundamentals $f=\frac{v}{2l}$ : $\frac{200}{1.2}=166.7$ , $\frac{200}{0.6}=333.3$ , $\frac{200}{0.4}=500\text{ Hz}$ .
Common frequency must be a whole multiple of each: LCM of $\frac{500}{3}, \frac{1000}{3}, 500$ is $1000\text{ Hz}$ .
So the answer is (b) 1000 Hz.

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