Simple  Harmonic Motion

 Given r=damping coefficient (dimension MT-1 ) s=stiffness constant (dimension MT-2) m=mass time=t A1 A2 2k=r/m w20= s/m (dimension T-2) k w0 ; relaxation  time tr =1/k=2m/r n=k / w0 n0=√(n2-1) w12=k2 - w20  ; w1= (ew1t +e-w1t)/2=coshw1t (ew1t -e-w1t)/2=sinhw1t A=(A1 +A2) B=(A1  - A2) (A-B)/(A+B) C1=(k+w1) / (k-w1) logC1 t1a=logC1/2w1 (time taken by mass to reach extreme position from mean position under boundary condition 1 ) x(t1a)=displacement in time t1a= A1e-kt1a(ew1t1a -e-w1t1a) velocity u at x=0 under boundary condition 1 =2A1w1 A1e-kt1a w1(ew1t1a +e-w1t1a) k(ew1t1a -e-w1t1a) coshw0n0t sinhw0n0t velocity at time t  =A1e-kt [w1(ew1t +e-w1t) - k(ew1t -e-w1t)]under boundary condition 1 velocity at time t  =(2nw0A1)(e-nw0t)[√(1-1/n2)*√(1+sinh2w1t) -sinhw1t] for n=1 or n >1  (boundary condition 1) displacement from mean position at time t =A1e-kt (ew1t -e-w1t)under boundary condition 1 displacement from mean position at time t =2A1 *e-nw0t sinh√(n2-1)*w0t  under boundary condition 1 acceleration at time t =-w20x-2kdx/dt  under boundary condition 1 amplitude part of acceleration at time t=-2A1w20e-nw0t  under boundary condition 1 frequency part of acceleration at time t(part 1)=sinhw0n0 t(1-2n2)under boundary condition 1 frequency part of acceleration at time t(part 2)=2n√(n2-1)coshn0w0t under boundary condition 1 acceleration at time t =-2A1w20e-nw0t [sinhw0n0 t(1-2n2) +2n√(n2-1)coshn0w0t] under boundary condition 1 (-Ak+Bw1) (-Aw1+Bk) tanhw1t1=(-Ak+Bw1) /(-Aw1+Bk) w1t1 C=[(A-B)/(A+B)] *(w1+k)/(w1-k) logC t1=logC/2w1 e-kt 1/e [Acoshw1t +Bsinhw1t] x=e-kt *[Acoshw1t +Bsinhw1t] x1=e-kt1 *[Acoshw1t1 +Bsinhw1t1] dx/dt =-kx+w1e-kt (Bcoshw1t +Asinhw1t) d2x/dt2 =-2kdx/dt +( w12-k2 )x At t=0, x=A, dx/dt=-kA+w1B, d2x/dt2 =A(w12+k2)-2Bkw1

 The differential equation of a damped Simple Harmonic Motion is given by  m*d2x/dt2 + r*dx/dt +sx =0 or d2x/dt2 + (r/m)dx/dt +(s/m)x =0 or  d2x/dt2 + 2k*dx/dt +w20 x =0 where r is damping coefficient , s is stiffness constant having dimension of surface tension, 2k=r/m  and  w20 = s/m, x is displacement from mean position. Solution is x=Aeαt  where A is amplitude and is a constant and  α is a constant. dx/dt=αx,  d2x/dt2 =α2x. Hence x* (α2 +2kα +w20)=0 or (α2 +2kα +w20)=0 α =-k ±√(k2  -w20)=-k ±w1 where √(k2  -w20)=w1   and general solution is x=A1*e[-k +√(k2  -w20)]t  +  A2*e[-k -√(k2  -w20)]t =e-kt [A1ew1t  +  A2*e-w1t ] dx/dt =w1e-kt [A1ew1t  -  A2*e-w1t] -ke-kt [A1ew1t  + A2*e-w1t] If we take x=Aeiwt   , the differential equn becomes x* (-w2 +2kiw +w20)=0 or w=ik ±i√(k2  -w20) and x=A1*e[-k +√(k2  -w20)]t  +  A2*e[-k -√(k2  -w20)]t =e-kt *[A1e[√(k2  -w20)]t  +  A2*e[-√(k2  -w20)]t ]=e-kt *[A1ew1t  +  A2*e-w1t ] So the solution is same as the previous one. The value of A1, A2 depend on Boundary conditions. Boundary condition 1: x=0 at t=0 i.e. the mass is at the mean position at t=0. Putting this condition, We get A1+A2=0 So x =e-kt [A1ew1t  -  A1*e-w1t ] =A1e-kt [ew1t  -  e-w1t ] or x=A1e-kt [ew1t  -  e-w1t ] dx/dt=w1A1e-kt [ew1t  +  e-w1t] - kA1e-kt [ew1t  -  e-w1t ] At t=0, x(0)=0 Velocity imparted at mean position =dx/dt (0)=2w1A1 To find out the time t1, the mass takes to reach amplitude A1, we put dx/dt =0 at A1 0=w1A1e-kt1 [ew1t1  +  e-w1t1] - kA1e-kt1 [ew1t1  -  e-w1t1 ] or w1[e2w1t1+1] =k[e2w1t1-1] or  t1=logC1/2w1       where C1=(k+w1)/(k-w1) Relaxation time=1/k=2m/r * Let A1 + A2=A, A1-A2=B, then A1=(A+B)/2, A2=(A-B)/2. Putting the values and rearranging, x=e-kt *[A (ew1t  + e-w1t)/2  + B (ew1t  - e-w1t)/2 ]    =  e-kt *[Acoshw1t +Bsinhw1t] x=e-kt *[Acoshw1t +Bsinhw1t] Here we assume that w1 > 0.......... case of over damping. dx/dt=(-Ak+Bw1)e-kt *coshw1t +(Aw1-Bk)e-kt *sinhw1t=e-kt*[(-Ak+Bw1)*coshw1t +(Aw1-Bk)*sinhw1t] Value of A,B determined by Boundary conditions. If we put the  condition , At t=0,x=0, then A=0 and x=Be-kt *sinhw1t=(ew1t  - e-w1t)/2*Be-kt dx/dt=(-k)Be-kt[(ew1t  - e-w1t)/2] +Be-kt[(ew1t  + e-w1t)/2]*w1 which  at t=0 equals Bw1 which means velocity v(0) at t=0 is equal to Bw1 which is the initial velocity imparted to the mass to enable it execute SHM. Either you fix initial velocity and find out B or fix B and find out initial velocity.  If we put t=0, and try to find  value of x, x=A at t=0. and dx/dt=(-Ak+Bw1) at t=0 * in The displacement in heavy damping, the amplitude decays exponentially with time  and the hyperbolic terms increase. The net effect at first is increase but soon the exponential term dominates and displacement slowly falls off to zero.What is the maximum displacement at time t1 ? at t=t1, dx/dt=0 or[(A-B)/(A+B)] *(w1+k)/(w1-k) = e2w1t ; or t1=logC/2w1 where C=[(A-B)/(A+B)] *(w1+k)/(w1-k) . When boundary conditions are such that A=0, C=(k+w1)/(k-w1) Critical damping : w1=0 x=e-kt *Acoshw1t=Ae-kt   Here we assume that w1 = 0.......... case of critical damping. When w1 is nearly equal to zero, w1=δ and x=A1eδt+A2e-δt=A1+A1δt+A2-A2δt=(A1+A2)+(A1-A2)δt=A+Bt Decay is purely exponential snd faster than over damping. The amplitude can , in general , decay exponentially with respect to time or space. In case of former, we call it relaxation time and for later, skin depth as in case of em waves in a conducting media. The criteria is that amplitude to fall to 1/e of its fixed value. For mechanical oscillator, relaxation time=1/k=2m/r. For skin depth=√(2/ωσμ) and has dimension of length.