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JEE Main 24 Feb 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 11
Total: 90
In the given figure, a mass
M
is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is
k
. The mass oscillates on a frictionless surface with time period
T
and amplitude
A
. When the mass is in equilibrium position as shown in the figure, another mass
m
is gently fixed upon it. The new amplitude of oscillation will be
24 Feb 2021 Shift1
A
√
M
+
m
M
A
√
M
M
+
m
A
√
M
−
m
M
A
√
M
M
−
m
Validate
Solution:
Given, initial amplitude
=
A
Velocity at mean position,
v
=
A
ω
Applying conservation of momentum at mean position, we get
M
1
v
1
=
M
2
v
2
M
A
ω
=
(
M
+
m
)
v
′
⇒
v
′
=
M
A
ω
M
+
m
=
M
A
√
k
M
M
+
m
v
′
=
A
′
ω
′
=
A
′
√
k
M
+
m
∴
A
′
=
M
A
√
k
M
M
+
m
×
√
M
+
m
k
A
′
=
√
M
M
+
m
A
© examsnet.com
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