A wire of length L , diameter ' d ' density of material ' Q ' is under tension ' T ', having fundamental frequency of vibration n A . Another wire of length 2 L , tensi

Question

A wire of length L , diameter ' d ' density of material ' Q ' is under tension ' T ', having fundamental frequency of vibration n A ​ . Another wire of length 2 L , tension 2 T , density 2 Q and diameter 3 d has fundamental frequency of vibration n B ​ . The ratio n B ​ : n A ​ is

TestHub · Accepted Answer

The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2 L} \frac{T}{m}$ , where $m$ is the linear mass density.

The linear mass density $m$ can be written in terms of diameter $d$ and density $Q$ as $m = π r^{2} Q = \frac{π d ^{2} Q}{4}$ .

Substituting $m$ in the formula for frequency, we get:

$n = \frac{1}{2 L} \frac{4 T}{π d ^{2} Q} = \frac{1}{L d} \frac{T}{π Q}$

For wire A:

$n_{A} = \frac{1}{L d} \frac{T}{π Q}$

For wire B, the given parameters are $L_{B} = 2 L$ , $T_{B} = 2 T$ , $Q_{B} = 2 Q$ , and $d_{B} = 3 d$ .

$n_{B} = \frac{1}{( 2 L ) ( 3 d )} \frac{2 T}{π ( 2 Q )} = \frac{1}{6 L d} \frac{T}{π Q}$

Taking the ratio of $n_{B}$ to $n_{A}$ :

$\frac{n _{B}}{n _{A}} = \frac{\frac{1}{6 L d} \frac{T}{π Q}}{\frac{1}{L d} \frac{T}{π Q}} = \frac{1}{6}$

The ratio $n_{B} : n_{A}$ is $1 : 6$ .

Answer: $1 : 6$

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