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Q.

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

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a

20N

b

10N

c

60N

d

40N

answer is C.

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Detailed Solution

Mass of bigger body M=4kgMass of smaller body m=1kgSmaller mass (m=1kg) executes S.H.M of  angular frequency ω=25rad/s Amplitude x=1.6cm=  1.6×10−2As we know,T=  2πmKOr, 2πω  = 2πmKOr,  125=1K[∵m=1kg;ω=25rad/s] Or,  K=625 Nm−1The maximum force exerted by the system on the floor=Mg+Kx+mg=  4×10+625×1.6×10−2+1×10=60N
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