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1)

The differential equation of the simple harmonic  motion given by x=Acos(nt+α) is 


A) d2xdt2n2x=0

B) d2xdt2+n2x=0

C) dxdtd2xdt=0

D) d2xdt2dxdt+nx=0

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)xnxtan(nt+α)]

                                                                                                   [From Eq.(ii)]

=[n2xsec2(nt+α)n2xtan2(nt+α)]

=n2x[sec2(nt+α)tan2(nt+α)]

 =n2x×1       [sec2θtan2θ=1]

     d2xdt2=n2x

    d2xdt2+n2x=0