Question

The total length of a sonometer wire fixed between two bridges is $110 \, cm$ . Now, two more bridges are placed to divide the length of the wire in the ratio $6:3:2$ . If the tension in the wire is $400 \, N$ and the mass per unit length of the wire is $0.01 \, kg \, m^{- 1}$ , then the minimum common frequency with which all the three parts can vibrate, is

• A. $1000Hz$
• B. $1100Hz$
• C. $100Hz$
• D. $110Hz$

Answer 1

Answer: (A)

$l_{1} : l_{2} : l_{3} \, \, \, = \, 6 : 3 : 2$
The frequancy in any $n^{t h}$ mode for a segment is $f=\frac{n v}{2 l}=constant$
no of loops $\propto$ length of the segment
so, no of loops are in the ratio 6:3:2
hence total loops =11
The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum
$\frac{\lambda }{2}=\frac{1}{11}\text{x}110=10cm$
$\text{f} = \frac{\text{V}}{\lambda } = \frac{1}{\lambda } \cdot \sqrt{\frac{\text{F}}{\mu }} = \frac{1}{\text{0.2}} \sqrt{\frac{4 0 0}{\text{0.01}}} = 1 0 0 0 Hz$