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JEE Advanced 2014 Paper 1
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© examsnet.com
Question : 4
Total: 60
One end of a taut string of length
3
m along the
x
axis is fixed at
x
=
0
. The speed of the waves in the string is
100
ms
−
1
. The other end of the string is vibrating in the
y
direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are)
y
(
t
)
=
A sin
π
x
6
cos
50
π
t
3
y
(
t
)
=
A sin
π
x
3
cos
100
π
t
3
y
(
t
)
=
A sin
5
π
x
6
cos
250
π
t
3
y
(
t
)
=
A sin
5
π
x
6
cos
250
π
t
Validate
Solution:
Taking y(t)=A f(x) g(t) & Applying the conditions:
1;here
x
=
3
m is antinode &
x
=
0
is node
2; possible frequencies are odd multiple of fundamental frequency.
where V
fundamental
=
v
4
l
=
25
3
Hz
The correct options are A,C,D
© examsnet.com
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