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 $π^{2} = 10$ .

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Solution

As system accelerate upward a pseudo force will act downward
$k x = m a x = \frac{4 \times 5}{1000} \times 100 c m x = 2 c m$
Amplitutde of SHM $= 2 c m$
$ω = \sqrt{\frac{K}{m}} = \sqrt{\frac{1000}{4}} = \sqrt{250}$
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
$y = 2 c m s i n [\sqrt{250} t + (- \frac{π}{2})] y = 2 c m c o s \sqrt{250} t$

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As system accelerate upward a pseudo force will act downwardkx=max=4×51000×100cmx=2cmAmplitutde of SHM=2cmω=√Km=√10004=√250Phase of the motion will lag through 90∘ because motion has started xcm before its mean position.So, equation of sum will bey=2cmsin[√250t+(−π2)]y=2cmcos√250t