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Byju's Answer
Standard XII
Physics
String Fixed at Both Ends
A wire of den...
Question
A wire of density
9
×
10
−
3
kg cm
−
3
is stretched between two clamps
1
m
apart. The resulting strain in the wire is
4.9
×
10
−
4
. The lowest frequency of the transverse vibrations in the wire (in
Hz
) is
(Young's modulus of wire
Y
=
9
×
10
10
Nm
−
2
)
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Solution
Fundamental frequency of vibration of the string,
f
=
1
2
l
√
T
μ
We can rewrite the above equation as,
f
=
1
2
l
√
T
ρ
A
f
=
1
2
l
√
Y
Δ
l
ρ
l
[Using
Y
=
T
/
A
Δ
l
/
l
]
From the data given,
f
=
1
2
×
1
√
9
×
10
10
×
4.9
×
10
−
4
9
×
10
−
3
×
10
6
=
35.00
Hz
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0
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String Fixed at Both Ends
Standard XII Physics
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