The correct option is A A→p,s;B→q,s;C→s;D→q
A: The graph of potential energy as function of displacement of a simple pendulum will be parabolic graph as given in option (p). The option (s) is also correct, if the mean position of the pendulum is at the origin.
Hence (A) → (p),(s)
B: The body is moving along positive x-axis, v>0; y is the displacement.
∴y=ut±12at2. For a=+ve and a=0, we get graphs which are satisfied by (q) and (s).
Hence (B)→(q),(s)
C: R=u2sin(2θ)g and R∝u2 for a fixed angle of projection and u=0,R=0
(C)→(s)
D: Time period of a simple pendulum is:
T=2π√lg or T2=4π2lg⇒y=4π2gx
(D)→(q)