AP EAPCET 07-Jul-2022 Shift 2 Paper

Given, length of wire =1.25m
strain produced =0.14%
density of wire, d=7.7×103kgm−3
young's modulus, Y=2.2×1011Nm−2
Fundamental frequency of vibration over a string is given by
f=‌

1

2L

√‌

T

µ

Where, T= Tension in string,
L= Length of string and
µ= mass per unit length of string.
So, we can express frequency asf=

1

2L

√

T∕A

M∕LA

From young's modulus of elasticity we have,
Y=

T∕A

∆l∕l

⇒

T

A

=22×1011×

0.14

100

⇒

T

A

=0.308×109
Substituting for
L,

T

A

and
d
in Eq.(i) we get, Fundamental frequency,
f=

1

2×1.25

×√

0.308×109

7.7×103

=

1

25

×√0.04×106
=

1

25

×√4×104=

2

25

×102=80‌Hz