Q.

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.
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a

1LTμ

b

52LTμ

c

12LTμ

d

32LTμ

answer is D.

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Detailed Solution

12LTμ=f

f1=n12LTμ,f2=n22LT16μ, here f1=f2n1=n24

For lowest possible frequency in the strings

n1=1,n2=4fmin=12LTμ

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