A vibrator makes 150 cm of a string to vibrate in 8 loops in the longitudinal arrangement when it is stretched by 150 N. The entire length of the string is then weighed and is found to weigh 400 mg. Then

Question

A vibrator makes 150 cm of a string to vibrate in 8 loops in the longitudinal arrangement when it is stretched by 150 N. The entire length of the string is then weighed and is found to weigh 400 mg. Then (A) frequency of the vibrator is 2 kHz (B) frequency of the vibrator is 1.5 kHz (C) distance between two nodes is 25 cm (D) distance between two nodes is 33 cm

Tardigrade Admin · Accepted Answer

f =( P /2 ℓ) √( T /μ), here P =8 ℓ=1.50 m , T =150 N, μ=( m /ℓ)=(400 × 10-6/1.50) f =(8/2 × 1.50) √(150 × 1.50/400 × 10-6) f =2000 Hz or 2 kHz

Q. A vibrator makes $150 c m$ of a string to vibrate in $8$ loops in the longitudinal arrangement when it is stretched by $150 N$ . The entire length of the string is then weighed and is found to weigh $400 m g$ . Then

frequency of the vibrator is 2 kHz

frequency of the vibrator is 1.5 kHz

distance between two nodes is 25 cm

distance between two nodes is 33 cm

$f = \frac{P}{2 ℓ} \frac{T}{μ}$ , here $P = 8$
$ℓ = 1.50 m, T = 150 N$ ,
$μ = \frac{m}{ℓ} = \frac{400 \times 1 0 ^{- 6}}{1.50}$
$f = \frac{8}{2 \times 1.50} \frac{150 \times 1.50}{400 \times 1 0 ^{- 6}}$
$f = 2000 Hz$ or $2 k Hz$

Q. A vibrator makes 150cm of a string to vibrate in 8 loops in the longitudinal arrangement when it is stretched by 150N. The entire length of the string is then weighed and is found to weigh 400mg. Then