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Tardigrade
Question
Physics
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g / m. When both the strings vibrate simultaneously the number of beats is :
Q. Each of the two strings of length
51.6
c
m
and
49.1
c
m
are tensioned separately by
20
N
force. Mass per unit length of both the strings is same and equal to
1
g
/
m
. When both the strings vibrate simultaneously the number of beats is :
3454
227
AIPMT
AIPMT 2009
Waves
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A
7
58%
B
8
16%
C
3
20%
D
5
7%
Solution:
l
1
=
0.516
m
,
l
2
=
0.491
m
,
T
=
20
N
Mass per unit length,
μ
=
0.001
k
g
/
m
Frequency,
v
1
=
2
×
0.516
1
0.001
20
v
2
2
×
0.491
1
0.001
20
∴
Number of beats
=
v
1
−
v
2
=
7.