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PhysicsMatrix match+8Question 40 of 60

IIT-JEE 2011 · Paper 2, Physics question 40

Column I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as λf\lambda_f. Match each system with statements given in Column II describing the nature and wavelength of the standing waves.

Column I

  • A
    Pipe closed at one end
    Pipe of length L closed at the end x = 0.
  • B
    Pipe open at both ends
    Pipe of length L open at both ends.
  • C
    Stretched wire clamped at both ends
    Wire of length L clamped at x = 0 and x = L.
  • D
    Stretched wire clamped at both ends and at mid-point
    Wire of length L clamped at both ends and at the mid-point L/2.

Column II

  • P
    Longitudinal waves
  • Q
    Transverse waves
  • R
    λf=L\lambda_f=L
  • S
    λf=2L\lambda_f=2L
  • T
    λf=4L\lambda_f=4L
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

Official answer: A → P, T; B → P, S; C → Q, S; D → Q, R

  • A Closed pipe: sound waves are longitudinal. The fundamental has a node at the closed end and an antinode at the open end: L=λ4L=\frac\lambda4, so λf=4L\lambda_f=4L. → P, T.
  • B Open pipe: antinodes at both ends, L=λ2L=\frac\lambda2, so λf=2L\lambda_f=2L. → P, S.
  • C String clamped at both ends: transverse waves, nodes at both ends, λf=2L\lambda_f=2L. → Q, S.
  • D Clamped also at the middle: a node at L/2L/2, so the lowest mode has λ2=L2\frac\lambda2=\frac L2 and λf=L\lambda_f=L. → Q, R.

Answer: A → P, T; B → P, S; C → Q, S; D → Q, R

Topics:Organ pipes and resonanceStanding waves in strings
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