A spring has natural length 40cm and spring constant 500N/m.A block of mass 1kg is attached at one end of the spring and other end of the spring is attached to ceiling. The block released from the position, where the spring has length 45cm, then
A
the block will perform SHM of amplitude 5cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
the block will have maximum velocity 30√5cm/sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
the block will have maximum acceleration 15m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
the minimum potential energy of the spring will be zero
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are B the block will have maximum acceleration 15m/s2 C the minimum potential energy of the spring will be zero D the block will have maximum velocity 30√5cm/sec From the image it is clear that, the amplitude of SHM will be a=3cm, also from the given data, ω=√km=10√5, thus maximum velocity vmax=aω=30√5cm−sec−1 Accelerationmax=aω2=15m−sec−2 Minimum potential energy is 0 at mean position.