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JEE Advanced 2013 Paper 2
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© examsnet.com
Question : 2
Total: 60
A particle of mass m is attached to one end of a mass-less spring of force constant k,lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t=0 with an initial velocity
u
0
. When the speed of the particle is
0.5
u
0
,it collies elastically with a rigid wall. After this collision :
the speed of the particle when it returns to its equilibrium position is
u
0
the time at which the particle passes through the equlibrium position for the first time is
t
=
Π
√
m
k
the time at which the maximum compression of the spring occurs is
t
=
4
Π
3
√
m
k
the time at which the particle passes throughout the equilibrium position for the second time is
t
=
5
Π
3
√
m
k
Validate
Solution:
v
=
μ
0
sin ωt (suppose
t
1
is the time of collision)
μ
0
2
=
μ
0
cos
ω
t
1
⇒
t
1
=
π
3
ω
Now the particle returns to equilibrium position at time
t
2
=
2
t
1
i.e
2
π
3
ω
with the same mechanical energy
i.e. its speed will
µ
0
.
Let
t
3
is the time at which the particle passes through the equilibrium position for the second time.
∴
t
3
=
T
2
+
2
t
1
=
π
ω
+
2
π
3
ω
=
5
π
3
ω
=
5
π
3
√
m
k
Energy of particle and spring remains conserved.
© examsnet.com
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