Question

The bridge of a sonometer is slightly displaced so that the length of the wire is decreased by $0.5\%$ and the tension in the wire is increased by $1\%$. The fundamental frequency of the wire

• A. increases by $1\%$
• B. decreases by $1\%$
• C. increases by $1.5\%$
• D. decreases by $1.5\%$

Answer 1

The correct option is A increases by $1\%$

Given : $\frac{l_1}{l_2}=\frac{100}{99.5}$
$\frac{T_1}{T_2}=\frac{100}{101}$
Fundamental frequency = $\sqrt{\frac{T}{ml}}$
So, $\frac{f_1}{f_2}=\sqrt{\frac{T_1}{m{l}_1}}\times \sqrt{\frac{m{l}_2}{T_2}}$
$=\sqrt{\frac{T_1}{T_2}}\times \sqrt{\frac{l_2}{l_1}}$
$=\sqrt{\frac{100}{101}}\times \sqrt{\frac{99.5}{100}}$
$=\sqrt{\frac{99.5}{101}}$
$=0.99$
$f_2=\frac{f_1}{0.99}=1.01f_1$
So, $f_2$ increases approximately by $1\%.$