Question

A $1\text{cm}$ long string vibrates with fundamental frequency of $256\text{Hz}$. If the length is reduced to $\frac{1}{4}\text{cm}$ keeping the tension unaltered, the new fundamental frequency will be (in Hz) (string is fixed at both ends)

• A. $64$
• B. $256$
• C. $512$
• D. $1024$

Answer 1

The correct option is D $1024$

Given,
$l=1\text{cm}=0.01\text{m}$
Fundamental frequency for string will be,
$f=\frac{v}{2l}=\frac{v}{2l}=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
For the given string ${}^{\prime }{\mu }^{\prime }$ will be fixed and tension is kept unchanged.
$\Rightarrow f\propto \frac{1}{l}$
$\Rightarrow \frac{f_1}{f_2}=\frac{l_2}{l_1}$
$\Rightarrow f_2=\left(f_1\right)\frac{l_1}{l_2}=\left(256\right)\times \frac{l_1}{\left(\frac{l_1}{4}\right)}$
$\therefore f_2=256\times 4=1024\text{Hz}$