Question

In a sonometer experiment, the fundamental frequency is $n$. If the distance between two bridges and the tension both are doubled from the original value, then the frequency will become

• A. $\sqrt{2}n$
• B. $2n$
• C. $\left(1/2\right)n$
• D. $\left(1/\sqrt{2}\right)n$

Answer 1

Answer: (D)

$\left(1/\sqrt{2}\right)n$

$n=\frac{1}{2l}\sqrt{\frac{T}{m}}$
where$,m=\text{Mass per unit length of sonometer wire},n=\text{Frequency of tuning fork},$
$l=\text{Vibrating length of the sonometer wire},T=\text{Tension in the wire}=Mg$

$\text{Now as per question new tension}=2T\text{and length}=2l$

$\text{Now,}{n}_1=\frac{1}{2\times 2l}\sqrt{\frac{2T}{m}}$
$\text{So dividing}{n}_1\text{by}n,\frac{n_1}{n},\text{we get},$
$\frac{n_1}{n}=\frac{1}{\sqrt{2}}$

$\text{So new frequency}{n}_1=\frac{1}{\sqrt{2}}n$