Answer:
Option B
Explanation:
Given ,x=Acos(nt+α) ...........(i)
On differentiating x w.r.t .'t' we get
dxdt=Addcos(nt+α)
=−Asin(nt+α)ddt(nt+α)
=−Asin(nt+α)(n)
⇒ dxdt=−Ansin(nt+α)
⇒ dxdt=−xnsin(nt+α)(cos(nt+α) [From Eq.(i)]
⇒ dxdt=−nxtan(nt+α) ........(ii)
Again differentiating w.r.t 't' , we get
d2xdt2=−[nxddttan(nt+α)+tan(nt+α)ddxnx]
=−[nxsec2(nt+α)ddt(nt+α)+tan(nt+α)dxdt×n]
=−[nxsec2(nt+α)n+ntan(nt+α)x−nxtan(nt+α)]
[From Eq.(ii)]
=−[n2xsec2(nt+α)−n2xtan2(nt+α)]
=−n2x[sec2(nt+α)−tan2(nt+α)]
=−n2x×1 [∵sec2θ−tan2θ=1]
⇒ d2xdt2=−n2x
⇒ d2xdt2+n2x=0