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JEE Main Physics Class 11 Oscillations Part 2 Questions
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© examsnet.com
Question : 71
Total: 76
Two identical springs of spring constant
2
k
are attached to a block of mass
m
and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is
[25 Feb 2021 Shift 2]
2
π
√
m
2
k
2
π
√
m
k
π
√
m
k
π
√
m
2
k
Validate
Solution:
Let spring constants of two springs be
k
1
and
k
2
.
Since, two springs are connected in parallel connection and parallel equivalent spring constant,
k
eq
=
k
1
+
k
2
⇒
k
e
q
=
2
k
+
2
k
=
4
k
As, time period,
T
=
2
π
√
m
k
eq
⇒
T
=
2
π
√
m
4
k
=
2
π
2
√
m
k
=
π
√
m
k
© examsnet.com
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