Under constant tension and constant mass per unit length, the note produced by a plucked string is \(500 Hz\) when the length of the string is \(0.90 m\). At what length is the frequency \(150 Hz\) ?
A. \(6 m\) B. \(3 m\) C. \(5 m\) D. \(4 m\)
Correct Answer: B
Explanation
Here, the correlating equation is \begin{aligned} &F=\frac{1}{2 l} \sqrt{\frac{T}{\mu}} \text { i.c. } F \alpha \frac{1}{I} \\ &F_{1} l_{1}=F_{2} l_{2} ; l_{2}=\frac{F_{1} l_{1}}{F_{2}} \\ &F_{1}=500 Hz ; l_{1}=0.90 m \\ &F_{2}=150 Hz , l_{2}=? \\ &l_{2}=\frac{500 \times 0.9}{150}=3 m \end{aligned}
