A body of mass m is hanging from a wire of length l. It is pulled by a distance Δl and left. It executes simple harmonic motion. If the young’s modulus of the wire is Y, area of cross-section is A, then the frequency of oscillation is
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
f=12πmlYA
b
f=12πYAlm
c
f=12πYAml
d
f=12πmYAl
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
For a body under stress we know that its Young’s modulus can be written as Y=FlAΔl. Here FA is stress and Δll is strain. This restoring force can be also written as F=−kΔl . Hence, we get k=FΔl=YAl. We know that for a simple harmonic motion the frequency of oscillation is given by f=12πkm . Substituting for the value of k we get f=12πYAml.