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