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Question
Physics
The maximum velocity of a particle executing SHM is 'v'. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value, the maximum velocity becomes :-
Q. The maximum velocity of a particle executing SHM is '
v
'. If the amplitude is doubled and the time period of oscillation decreased to
1/3
of its original value, the maximum velocity becomes :-
2352
227
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A
18v
0%
B
12v
0%
C
6v
100%
D
3v
0%
Solution:
V
m
a
x
1
=
ω
A
=
(
T
2
π
)
A
=
V
V
m
a
x
2
=
(
T
/3
2
π
)
(
2
A
)
=
6
(
T
2
π
)
A
=
6
V