Find the distance between two consecutive nodes if a string has tension 0.25 N, mass per unit length is 1 mg/m and frequency is 250 Hz.

Question

Find the distance between two consecutive nodes if a string has tension 0.25 N, mass per unit length is 1 mg/m and frequency is 250 Hz. (A) 2 m (B) 1 m (C) 3 m (D) 1.5 m

Tardigrade Admin · Accepted Answer

The speed of transverse wave on the string will be, v=√(T/m) v=√(0.25/1×10-6) v = 500 m/sec and v = f λ So, λ=(500/250) λ=2 m Distance between two consecutive nodes is given by, =(λ/2)=1 m

Q. Find the distance between two consecutive nodes if a string has tension $0.25 N$ , mass per unit length is $1 m g / m$ and frequency is $250 Hz$ .

2 m

1 m

3 m

1.5 m

The speed of transverse wave on the string will
be, $v = \frac{T}{m}$
$v = \frac{0.25}{1 \times 1 0 ^{- 6}}$
$v = 500 m / sec$
and
$v = f λ$
So, $λ = \frac{500}{250}$
$λ = 2 m$
Distance between two consecutive nodes is given by,
$= \frac{λ}{2} = 1 m$

Q. Find the distance between two consecutive nodes if a string has tension 0.25N, mass per unit length is 1mg/m and frequency is 250Hz.

2 m

1 m

3 m

1.5 m

The speed of transverse wave on the string will be, v=mT​​ v=1×10−60.25​​ v=500m/sec and v=fλ So, λ=250500​ λ=2m Distance between two consecutive nodes is given by, =2λ​=1m

Q. Find the distance between two consecutive nodes if a string has tension $0.25 N$ , mass per unit length is $1 m g / m$ and frequency is $250 Hz$ .

The speed of transverse wave on the string will
be, $v = \frac{T}{m}$
$v = \frac{0.25}{1 \times 1 0 ^{- 6}}$
$v = 500 m / sec$
and
$v = f λ$
So, $λ = \frac{500}{250}$
$λ = 2 m$
Distance between two consecutive nodes is given by,
$= \frac{λ}{2} = 1 m$