When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2.0 kg block to the spring and if the block is pulled through 10 cm and released the maximum velocity in it in m/s is : (Acceleration due to gravity =10m/s2)
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a
0.5
b
1
c
2
d
4
answer is B.
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Detailed Solution
Initially when 1 kg mass is suspended then by using F=kx⇒mg=kx⇒k=mgx=1×105×10−2=200Nm Further, the angular frequency of oscillation of 2 kg mass is ω=kM=2002=10 rad/sec Hence, vmax=aω=(10×10−2)×10=1 m/s