SHMVery Hard
Question
A system is shown in the figure. The time period for small oscillations of the two blocks will be:-
Options
A.
B.
C.
D.
Solution
Both the spring are in series
∴ Keq =
Time period T =
where μ =
Here = ∴
∴ T =
OR
Total extension = 2x
By energy conversation
E =
Keq(2x)2 + 
mv2
E =
ax2 + 
mv2 + 
mv2 = 
kx2 + mv2
there is no loss of energy
= 0 ⇒ 
kxv + 2mva = 0 ⇒ 
= - 2mva
a = -
⇒ - ω2x = - 
⇒ ω = 
T =
∴ Keq =
Time period T =
where μ =
Here = ∴
∴ T =
OR
Total extension = 2x
By energy conversation
E =
Keq(2x)2 + mv2E =
ax2 + mv2 + mv2 = kx2 + mv2there is no loss of energy
= 0 ⇒ kxv + 2mva = 0 ⇒ = - 2mvaa = -
⇒ - ω2x = - ⇒ ω = T =
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