KEAM 2018 Physics and Chemistry Paper

Given, ‌‌a∝x(t)....(i)
According option (a),
x(t)=sin‌ω‌t,ω>0
and a
or a=−ω2x(t)....(ii)
According option (b),
x(t)=sin‌ω‌t+cos‌ω‌t,ω>0
andor a=−(ω2‌sin‌ω‌t+ω2‌cos‌ω‌t) or a=−ω2x(t) ....(iii)
According option (c),
x(t)=eωt,ω>0
∴‌‌v(t)=‌

dx(t)

dt

=ωeωt
and a=‌

dv(t)

dt

=ω2eωt
Or a=ω2x(t).....(iv)
According option (d),
x(t)=eωt+sin‌ω‌t‌ω>0
∴‌‌v(t)=‌

dx(t)

dt

=ωeωt+ω‌cos‌ω‌t
and a=‌

dv(t)

dt

=ω2eωt−ω2‌sin‌ω‌t
or a=ω2(eωt−sin‌ω‌t).....(v)
According option (e),
x(t)=eω1t+e−ω2t,ω1 and ω2>0
∴‌‌v(t)=‌

dx(t)

dt

=ω1eω1t−ω2e−ω2t
and ‌‌a=‌

dv(t)

dt

=ω12eω1t+ω22e−ω2t
or ‌‌a=ω12eω1t+ω22e−ω2t ....(vi)
Hence by comparing Eqs. (ii), (iii), (iv), (v) and (vi) with Eq. (i), option (c) is correct.