Question

The fundamental frequency of a sonometer wire increases by 5 Hz if its tension is increased by 21%. Instead if the length of wire is decreased by 10%, then the fundamental frequency becomes

• A. 45.45 Hz
• B. 50 Hz
• C. 55.55 Hz
• D. None of the above

Answer 1

The correct option is C 55.55 Hz

Fundamental frequency of sonometer is given by
$F_0=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
When the tension increased by 21%
$F_0+5=\frac{1}{2l}\sqrt{\frac{1.21T}{\mu }}=1.1F_0$
$F_0+5=1.1F_0$
$F_0=\frac{5}{0.1}=50Hz$
When the length of wire decreased by 10%
Now, $F^{\prime }=\frac{1}{2\times 0.9l}\sqrt{\frac{T}{\mu }}$
$F^{\prime }=\frac{1}{0.9}{F}_0=\frac{10}{9}\times 50=\frac{500}{9}$
= 55.55 Hz.