Question

A massless spring is hanging vertically from the ceiling. A mass m is attached to the bottom end of the spring and released from rest. How close to its final resting position is the mass when βt = 1 given that the mass finally comes to rest a distance d below the point from which it was released and the oscillator is critically damped.

Answer 1

At the final rest position, the weight of the mass is exactly balanced by the Restoring force of the spring.

Now the weight of mass $W= mg$

and the magnitude of restoring force at maximum displacement is $|F_{s}|= kd$

then equating above two, we get

$mg= kd$............................(1)

Alo we know that the natural frequency of the oscillation is $\omega_o=\sqrt{\frac{k}{m}}$

or using equation 1, $\omega_o=\sqrt{\frac{g}{d}}$ ...................(2)

If y(t) denotes the displacement from rest. then for critical damping we know that displacement is given by

  $y(t)= (C_1+C_2t)e^{-\omega _o t}$...........................(3)

Now taking positive y in downward direction thus displacement at t=0s is

$y(0)= -d$

and since it started from rest, thus initial velocity will be $\dot{y}(0)= 0$

Now putting t=0 in equation 3, we get

$y(0)= (C_1+C_2(0))e^{-\omega _o 0}= (C_1)(1)$

$C_1= -d$.....................(4)

Similarly, we'll get after putting the initial conditions,

  $\dot{y}(t)= (C_2)e^{-\omega _o t}+(C_1+C_2t)e^{-\omega _o t}(-\omega _o)$

  $\dot{y}(0)=0= (C_2)(1)+(-d+0)(1)(-\omega _o)$

  $C_2= -d\omega_o$......................(5)

Putting above two equations(4 and 5) in equation 3, we get

$y(t)= (C_1+C_2t)e^{-\omega _o t}$

$y(t)= (-d+(-d\omega_o)t)e^{-\omega _o t}$

$y(t)= -d(1+\omega_o t)e^{-\omega _o t}$

Then putting time, t=1 in above we'll get displacement

  $y(1)= -d(1+\omega_o (1))e^{-\omega _o (1)}$

  $y(1)= -d(1+\omega_o)e^{-\omega _o}$

Now using equation 2 in above,

  $y(1)= -d(1+\sqrt{\frac{g}{d}})e^{-\sqrt{\frac{g}{d}}}$( ANS)