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Question: Answered & Verified by Expert
A light string is tied at one end to a fixed support and to a heavy string of equal length L at the other end A as shown in the figure ( Total length of both strings combined is 2L). A block of mass M is tied to the free end of heavy string. Mass per unit length of the strings are μ and 16μ and tension is T. Find lowest positive value of frequency such that junction point A is a node.

PhysicsWaves and SoundJEE Main
Options:
  • A 1LTμ
  • B 52LTμ
  • C 32LTμ
  • D 12LTμ
Solution:
2077 Upvotes Verified Answer
The correct answer is: 12LTμ
f1=n12LTμf2=n22LT16μ

f1=f2n1=n24

n1=1n2=4

fmin=12LTμ

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