A wire of length L and mass per unit length 6.0 × 10-3 kgm-1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is :

Question

A wire of length L and mass per unit length 6.0 × 10-3 kgm-1 is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is : (A) 8.1 m (B) 2.1 m (C) 1.1 m (D) 5.1 m

Tardigrade Admin · Accepted Answer

(nv/2ℓ) = 420 ((n+1)v/2ℓ) = 490 (v/2ℓ) = 70 ℓ = (v/140) = (1/140)√(540/6×10-3) = (1/140)√90×103 ℓ = (300/140) = 2.142

Q. A wire of length $L$ and mass per unit length $6.0 \times 1 0^{- 3} k g m^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are: $420 Hz$ and $490 Hz$ . Then $L$ in meters is :

$8.1 m$

$2.1 m$

$1.1 m$

$5.1 m$

$\frac{n v}{2 ℓ} = 420$
$\frac{( n + 1 ) v}{2 ℓ} = 490$
$\frac{v}{2 ℓ} = 70$
$ℓ = \frac{v}{140} = \frac{1}{140} \frac{540}{6 \times 1 0 ^{- 3}} = \frac{1}{140} 90 \times 1 0^{3}$
$ℓ = \frac{300}{140} = 2.142$

Q. A wire of length L and mass per unit length 6.0×10−3kgm−1 is put under tension of 540N. Two consecutive frequencies that it resonates at are: 420Hz and 490Hz. Then L in meters is :

8.1m

2.1m

1.1m

5.1m