If n1, n2 and n3, are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
1n = 1n1+1n2+1n3
b
1n = 1n1+1n2+1n3
c
n = n1+n2+n3
d
n = n1+n2+n3
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Total length of string l = l1+l2+l3(As string is divided into three segments)n1 = 12l1Tμ; n2 = 12l2Tμ; n3 = 12l3TμHence frequency ∝ 1length(∵ n = 12lTm)∴ 1n= 2lTμ = 2l1Tμ+2l2Tμ+2l3Tμ = 1n1+1n2+1n3