Physics Circle home Student Login

Home›Chapters›Wave Motion PYQs›Stationary Waves & Vibrating Strings›Question 115

NEET 1995 (AIPMT) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from NEET 1995 (AIPMT)

Q115NEET 1995 · AIPMTEasy

Stationary Waves & Vibrating Strings

The length of a sonometer wire ABAB is 110 cm110\text{ cm}. Where should the two bridges be placed from AA to divide the wire in 3 segments whose fundamental frequencies are in the ratio of 1:2:31:2:3?

Show answer and solution

Answer: (a)

  1. f∝1lf \propto \dfrac{1}{l}: l1:l2:l3=1:12:13=6:3:2l_1 : l_2 : l_3 = 1 : \dfrac{1}{2} : \dfrac{1}{3} = 6 : 3 : 2.
  2. Each part =11011=10 cm⇒60,30,20 cm= \dfrac{110}{11} = 10\text{ cm} \Rightarrow 60, 30, 20\text{ cm}.
  3. Bridges from A at 60 cm60\text{ cm} and 90 cm90\text{ cm}.
  4. So the answer is (a) 60 cm60\text{ cm} and 90 cm90\text{ cm}.
Report a mistake

More questions on Stationary Waves & Vibrating Strings