Question

A wire of length $L$ and mass $6\times {10}^{-3}kg{m}^{-1}$ per unit length is put under the tension of $540N$. Two consecutive frequencies that it resonates at are: $420Hz$ and $490Hz$ . Then $L$ in meter is

• A. $8.1m$
• B. $2.1m$
• C. $1.1m$
• D. $5.1m$

Answer 1

The correct option is B

$2.1m$

Step 1. Given data

Frequencies, $f_1=420Hz$

$f_2=490Hz$

Tension, $T=540N$

Mass per unit length, $u=6.6\times {10}^{-3}kg{m}^{-1}$

Step 2. Finding the length, $L$

Fundamental frequency$=f_2-f_1$

$=490-420$

$=70Hz$

By using the formula of the fundamental frequency, $f$

$f=\frac{1}{2L}\sqrt{\frac{T}{u}}$

$70=\frac{1}{2L}\sqrt{\frac{540}{6\times {10}^{-3}}}$

$L=2.1m$

Hence, the correct option is $\left(B\right)$.