Question

The total length of a sonometer wire fixed between two bridges is $110cm$ . 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 $400N$ and the mass per unit length of the wire is $0.01kg{m}^{-1}$ , 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 $n^{th}$ mode for a segment is $f=\frac{nv}{2l}=constant$
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
$\frac{\lambda }{2}=\frac{1}{11}\text{x}110=10cm$
$\text{f}=\frac{\text{V}}{\lambda }=\frac{1}{\lambda }\cdot \sqrt{\frac{\text{F}}{\mu }}=\frac{1}{\text{0.2}}\sqrt{\frac{400}{\text{0.01}}}=1000Hz$