Comment
Share
Q)

# Two bodies of masses 1 kg and 4 kg are connected to a vertical spring , as shown in the figure . The smaller mass executes simple harmonic motion of angular frequency 25 rad/s , and amplitude 1.6 cm while the bigger mass remains stationary on the ground . The maximum force exerted by the system on the floor is (take g=$\;10 ms^{-2}$) .

$(a)\;20\;N\qquad(b)\;10\;N\qquad(c)\;60\;N\qquad(d)\;40\;N$

Comment
A)

when the smaller block is at the lower extreme from the equilibrium position it has an upward acceleration.as we know that in shm acceleartion is maximum at extreme position.

we can calculate acceleration at extreme position with the formula

a=Aw^2

a=1.6*10^-2*(625)

=10

so there is upward force  of 10N

So the net downward force is 40-10=30N

so it is the minimum force .

If the ball is some where other than extreme position it does apply any force on the block.at that time the down awrd force is due the bigger blocks own weight i.e 40N