Question 279: Waves

Question

A wire of length L , diameter ' d ' density of material ' e ' is under tension ' T ', having fundamental frequency of vibration $\mathrm{n}_{\mathrm{A}}$. Another wire of length 2 L , tension 2 T , density 2 e and diameter 3 d has fundamental frequency of vibration $\mathrm{n}_{\mathrm{B}}$. The ratio $\mathrm{n}_{\mathrm{B}}: \mathrm{n}_{\mathrm{A}}$ is

Accepted Answer

The fundamental frequency of vibration of a stretched string is given by:
$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $
where
$L$ = length of the string,
$T$ = tension,
$\mu$ = mass per unit length.
Step 1: Find $\mu$ in terms of density and diameter
The mass per unit length, $\mu = 	ext{density} 	imes 	ext{area of cross-section}$.
For a wire of diameter $d$:
$ 	ext{Area} = \frac{\pi d^2}{4} $
So,
$ \mu = e 	imes \frac{\pi d^2}{4} $
Step 2: Write expressions for $n_A$ and $n_B$
For wire A:
Length = $L$
Tension = $T$
Density = $e$
Diameter = $d$
So,
$ \mu_A = e \frac{\pi d^2}{4} $
$ n_A = \frac{1}{2L} \sqrt{\frac{T}{e \frac{\pi d^2}{4}}} $
For wire B:
Length = $2L$
Tension = $2T$
Density = $2e$
Diameter = $3d$
So,
$ \mu_B = 2e 	imes \frac{\pi (3d)^2}{4} = 2e 	imes \frac{\pi 	imes 9d^2}{4} = \frac{18e\pi d^2}{4} $
$ n_B = \frac{1}{2(2L)} \sqrt{\frac{2T}{\frac{18e\pi d^2}{4}}} $
Step 3: Simplify $n_A$ and $n_B$
$n_A$:
$ n_A = \frac{1}{2L} \sqrt{\frac{T}{e \frac{\pi d^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{e\pi d^2}} $
$n_B$:
$ n_B = \frac{1}{4L} \sqrt{\frac{2T}{\frac{18e\pi d^2}{4}}} = \frac{1}{4L} \sqrt{\frac{2T 	imes 4}{18e\pi d^2}} = \frac{1}{4L} \sqrt{\frac{8T}{18e\pi d^2}} = \frac{1}{4L} \sqrt{\frac{4T}{9e\pi d^2}} $
Step 4: Find the ratio $\frac{n_B}{n_A}$
Take the simplified forms:
$ n_A = \frac{1}{2L} \sqrt{\frac{4T}{e\pi d^2}} $
$ n_B = \frac{1}{4L} \sqrt{\frac{4T}{9e\pi d^2}} $
So,
$ \frac{n_B}{n_A} = \frac{\frac{1}{4L} \sqrt{\frac{4T}{9e\pi d^2}}}{\frac{1}{2L} \sqrt{\frac{4T}{e\pi d^2}}} $
$ = \frac{1}{4L} \div \frac{1}{2L} 	imes \frac{\sqrt{\frac{4T}{9e\pi d^2}}}{\sqrt{\frac{4T}{e\pi d^2}}} $
$ = \frac{1}{2} 	imes \frac{\sqrt{\frac{4T}{9e\pi d^2}}}{\sqrt{\frac{4T}{e\pi d^2}}} $
Let's write the square roots explicitly:
$ \frac{\sqrt{\frac{4T}{9e\pi d^2}}}{\sqrt{\frac{4T}{e\pi d^2}}} = \sqrt{\frac{\frac{4T}{9e\pi d^2}}{\frac{4T}{e\pi d^2}}} = \sqrt{\frac{4T}{9e\pi d^2} 	imes \frac{e\pi d^2}{4T}} = \sqrt{\frac{1}{9}} = \frac{1}{3} $
So:
$ \frac{n_B}{n_A} = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6} $
Final Ratio:
$ n_B : n_A = 1 : 6 $
Correct option: Option C

Answer

$1: 2$

Answer

$1: 4$

Answer

$1: 6$

Answer

$1: 8$

Waves

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