Question

A vibrator makes 150 cm of a string to vibrate
in 6 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 3 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​

Answer 1

The distance between two nodes is 25cm.

Length of the string = 150cm (GIven)

Loops = 6  (GIven)

Force= 150N  (GIven)

Velocity of wave = V = √ T/u, where T is the tension and u is the mass of string.

Therefore, V =750m/s

As the string is having 6 loops thus,

L =6λ/2

= λ = 50

Frequency = V/λ

= 750/0.5

= 1500 Hz

Distance between two nodes = λ/2  

= 50/2

= 25

Thus, the distance between two nodes is 25cm.