Question

A spring block (force constant k=1000 N/m and mass m=4 kg) system is suspended from the ceiling of an elevaror such that block is initially at rest.The elevator begins to move upwards at t=0.Acceleration time graph of the elevator is shown in the figure. Draw the displacement x (from its initial position taking upwards as positive) vs time graph of the block with respect to the elevator starting from t=0 to t=1 sec.Take ${\pi }^2=10$.

Answer 1

As system accelerate upward a pseudo force will act downward

$kx=max=\frac{4\times 5}{1000}\times 100cmx=2cm$

Amplitutde of SHM$=2cm$

$\omega =\sqrt{\frac{K}{m}}=\sqrt{\frac{1000}{4}}=\sqrt{250}$

Phase of the motion will lag through ${90}^{\circ }$ because motion has started $x$cm before its mean position.

So, equation of sum will be

$y=2cmsin[\sqrt{250}t+(-\frac{\pi }{2})]y=2cmcos\sqrt{250}t$