Question

The fundamental frequency of a sonometer wire increases by $6Hz$ if its tension is increased by $44\%$,
keeping the length constant. Find the change in the fundamental frequency (in Hz ) of the sonometer wire when the length of the wire is increased by $20\%$, keeping the original tension in the wire.

Answer 1

As we know, fundamental frequency of sonometer wire,
$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$f\alpha \sqrt{T}$(if l,$\mu$ are $\text{constant}$
Let the fundamental frequency of the sonometer wire be $f$.
When tension is increased by $44\%$, frequency of the wire increased by $6Hz$.
$\Rightarrow \frac{f}{f+6}=\sqrt{\frac{T}{T(1+\frac{44}{100})}}$
$\frac{f}{f+6}=\sqrt{\frac{1}{1.44}}=\frac{1}{1.2}$
$1.2f=f+6$

$\Rightarrow f=30Hz$

As we know, fundamental frequency of sonometer wire,
$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$f\alpha \frac{1}{l}$(if $T,\mu$ are constant)
when the length of the sonometer wire is increased by$20\%$
$fl=f^{\prime }{l}^{\prime }$
$fl=f^{\prime }l=f^{\prime }l(1+\frac{20}{100})$
$30l=f^{\prime }l\times 1.2$
$\Rightarrow f^{\prime }=25Hz$
Change in fundamental frequency
$\Delta f=f^{\prime }-f$
$=25-30=-5HZ$

$|\Delta f|=5Hz$