Question

The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by 8N, it is in resonance with first overtone of the closed tube. The initial tension in the wire is

• A. 16 N
• B. 8 N
• C. 4 N
• D. 1 N

Answer 1

The correct option is D 1 N

By laws of transverse vibration of strings:

$\nu \propto \sqrt{T}$

Let $T$ be the initial tension:

So, ${\nu }_c\propto \sqrt{T}$

So, $\frac{{\nu }_c}{3{\nu }_c}=\frac{\sqrt{T}}{\sqrt{T+8}}$ ( $\because$ first overtone $=3{\nu }_c$ )

$\frac{1}{9}=\frac{T}{T+8}$

$T+8=9T$

$8=8T$

$\Rightarrow T=1N$