Q. A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. the string is set into vibration using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string
A) 5
B) 6
C) 2
D) 3/2
Explanation
(a) Distance between the successive nodes,
$d = \frac{\lambda}{2}$
$= \frac{v}{2f}$
$= \frac{\sqrt{T/\mu}}{2f}$
Substituting the value we get
D=5cm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.