JEE Main | 2014SHMHard
Question
A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a in same direction then
Options
A.Amplitude of motion is 3a
B.Time period of oscillations is 8τ
C.Amplitude of motion is 4a
D.Time period of oscillations is 6τ
Solution
As it starts from rest, we have
x = Acosωt. At t = 0, x = A
when t = τ, x = A - a
when t = 2τ, x = A - 3a
⇒ A - a = Acosωτ
A - 3a = Acos2ωτ
As cos2ωτ = 2cos2ωτ - 1
⇒
= 2 
-1
A2 - 3aA = A2 + 2a2 - 4Aa
a2 = 2aA
A = 2a
Now, A - a = Acosωτ
⇒ cos ωτ =
=
⇒ T = 6τ
x = Acosωt. At t = 0, x = A
when t = τ, x = A - a
when t = 2τ, x = A - 3a
⇒ A - a = Acosωτ
A - 3a = Acos2ωτ
As cos2ωτ = 2cos2ωτ - 1
⇒
= 2 -1 A2 - 3aA = A2 + 2a2 - 4Aa
a2 = 2aA
A = 2a
Now, A - a = Acosωτ
⇒ cos ωτ =
=⇒ T = 6τ
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