Question

A string is stretched between fixed points separated by $75.0cm$. It is observed to have resonant frequencies of $420Hz$ and $315Hz$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is :

• A. $105Hz$
• B. $1.05Hz$
• C. $1005Hz$
• D. $10.5Hz$

Answer 1

The correct option is A $105Hz$

The string fixed at both end, frequency is

$f=315=\frac{nv}{2l}$. . . . .(1)

Next frequency,

$420=\frac{(n+1)v}{2l}$. . . .(2)

Dividing equation (2) by (1), we get

$\frac{420}{315}=\frac{n+1}{n}$

$420n=315n+315$

$n=3$

Put the value of n in equation (1)

$\frac{v}{2l}=105$. . . . .(3)

The lowest value of $n=1$

The lowest frequency is, $f_0=\frac{nv}{2l}$

$f_0=\frac{\left(1\right)v}{2l}$ ( from equation 3)

$f_0=105Hz$

The correct option is A.