JEE Main 27-Aug-2021 Shift 1 Solved Paper

It is given that motion is free simple harmonic motion.
Consider a spring mass system performing free simple harmonic motion.
Displacement of mass is given as
x=x0‌sin‌ω‌t
The potential energy stored in spring for displacement x can be given as
PE=

1

2

kx2
Substituting the value of x, we get
PE=

1

2

kx02sin2ωt
Here, k is spring constant and x0 is also constant.
PE=Csin2ωt (∴C=

1

2

kx02)
Simplify the value of sin2 ωt, we get
PE=C[

1−cos‌2‌ω‌t

2

]
The value of function cos2ωt will always lies between −1 and 1.
Thus, the value of potential energy will be greater than or equal to zero. 0 ≤ PE
This condition is only satisfies by graph (d).