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JEE Main 2017 Solved Paper Solved Paper
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© examsnet.com
Question : 1
Total: 90
A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like
Validate
Solution:
Time taken to reach the extreme position from equilibrium position is
T
4
.Velocity is maximum at equilibrium position and zero at extreme position.
V
=
A
ω
c
o
s
ω
t
K.E
=
1
2
m
v
2
(m is the mass of particle and v is the velocity of particle)
K.E.
=
1
2
m
A
2
ω
2
c
o
s
2
ω
t
Hence graph of K.E. v/s time is square cos function
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