UPSEE 2018 Solved Paper

According to law of conservation of energy, the total energy possessed by the particle should be same everywhere.
Let origin be x=0.
∴TE(x=0)=TE(x=a)
i.e. PE(x=0)+KE(x=0)=PE(x=a)+KE(x=a)

K(0)2

2

+Ae−α×02+KE(x=0)
=Ka22+Ae−αa2+

1

2

mva2
i.e. A+KE(x=0)=

ka2

2

+Ae−αa2+

1

2

mva2
So, A≤

ka2

2

+Ae−αa2+

1

2

mva2 [∵KE(x=0). is unknown, A<RHS],
if KE(x=0)=0,A=RHS]
i.e. A(1−e−αa2)≤

ka2+mva2

2

i.e. A≤

ka2+mva2

2(1−e−αa2)