AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Given, diameter, d1=0.9mm=0.9×10−3m
=9×10−4m
d2=0.93mm=9.3×10−4m
Frequency of fundamental tone is given as
f=‌

1

2l

√‌

T

m

....(i)
where, m= mass per unit length. =‌

M

l

=‌

V⋅ρ

l

=‌

A⋅l

l

⋅ρ=Aρ
⇒‌‌m=‌

πd2ρ

4

...(ii)
From Eqs. (i) and (ii), we get
⇒f=‌

1

Id

√‌

T

πρ

⇒f∝‌

1

d

∴ Percentage change in frequency
=‌

f2−f1

f1

×100=(‌

f2

f1

−1)×100 =(‌

7

4.3

−1)×100=−3.2%