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JEE Main Physics Class 11 Waves Part 1 Questions
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© examsnet.com
Question : 90
Total: 100
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
[2006]
105 Hz
1.05 Hz
1050 Hz
10.5 Hz
Validate
Solution:
It is given that
315
H
z
and
420
H
z
are two resonant frequencies, let these be
n
th
and
(
n
+
1
)
th
harmonies, then
we have
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
H
z
The lowest resonant frequency is when
n
=
1
Therefore lowest resonant frequency
=
105
H
z
© examsnet.com
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