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AIEEE 2006 Solved Paper
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© examsnet.com
Question : 76
Total: 143
A string is stretched between fixed points separated by
75.0
cm
. It is observed to have resonant frequencies of
420
Hz
and
315
Hz
. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
105
Hz
1.05
Hz
1050
Hz
10.5
Hz
Validate
Solution:
Given
n
v
2
ℓ
=
315
and
(
n
+
1
)
v
2
ℓ
=
420
⇒
n
+
1
n
=
420
315
⇒
n
=
3
Hence
3
×
v
2
ℓ
=
315
⇒
v
2
ℓ
=
105
Hz
Lowest resonant frequency is when
n
=
1
Therefore lowest resonant frequency
=
105
Hz
.
© examsnet.com
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