Question

A wire of length $l$ having tension $T$ and radius $r$ vibrates with natural frequency $f$. Another wire of same metal with length $2l$ having tension $2T$ and radius $2r$ will vibrate with natural frequency

• A. $f$
  
• B. $2f$
  
• C. $22f$
  
• D. $\frac{f}{\sqrt{2}}$
  

Answer 1

The correct option is D $\frac{f}{\sqrt{2}}$

Initially length $=l$, Tension $=T$ & Radius $=r$

Natural frequency $F=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Another wire have, length $=2l$, Tension $=T$ & Radius $=2r$

$\therefore$ It's natural frequency $F^1=\frac{1}{2\times 2l}\sqrt{\frac{2T}{\mu }}$

$=\frac{\sqrt{2}}{2}\times \frac{1}{2l}\sqrt{\frac{T}{\mu }}=\frac{1}{\sqrt{2}}F$

Hence $F/\sqrt{2}$ will be the correct option.

$\therefore$ Option (D) is correct.