A wire of density 9 10^3 kg /m^3 is stretched between two clamps … - Vidyadip

MCQ

A wire of density $9 \times 10^3 kg /m^3$ is stretched between two clamps $1 m$ apart and is subjected to an extension of $4.9 \times 10^{-4} m$. The lowest frequency of transverse vibration in the wire is ..... $Hz$ ($Y = 9 \times 10^{10} N / m^2$)

A
$40$
✓
$35 $
C
$30$
D
$25$

✓

Answer

Correct option: B.

$35 $

b
(b) For wire if

$M =$ mass, $\rho =$ density, $A = $Area of cross section

$V = $ volume, $l =$ length, $\Delta l$ = change in length

Then mass per unit length $m = \frac{M}{l} = \frac{{Al\rho }}{l} = A\rho $

And Young’s modules of elasticity

$y = \frac{{T/A}}{{\Delta l/l}}$

==> $T = \frac{{Y\Delta lA}}{l}$.

Hence lowest frequency of vibration $n = \frac{1}{{2l}}\sqrt {\frac{T}{m}} $

$ = \frac{1}{{2l}}\sqrt {\frac{{y\left( {\frac{{\Delta l}}{l}} \right)A}}{{A\rho }}} = \frac{1}{{2l}}\sqrt {\frac{{y\Delta l}}{{l\rho }}} $

==> $n = \frac{1}{{2 \times 1}}\sqrt {\frac{{9 \times {{10}^{10}} \times 4.9 \times {{10}^{ - 4}}}}{{1 \times 9 \times {{10}^3}}}} = 35Hz$

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