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A block of mass 400gm is hung vertically with the help of a massless spring.Initially block is in equillibrium position.Spring cinstant of the spring is k = 200N/m.Natural length of the spring is 40cm.Now the block is streched to some distance such that the total length of the spring becomes 45cm and then released.Qn) Amplititude of simple harmonic motion of the block is..........Pls explain....

vishwajit pureti , 13 Years ago
Grade 12
anser 1 Answers
Amitabh Shrivastava

Last Activity: 13 Years ago

Amplitude==45-initial length.

to find initial length,

force=m*g=k*x=>0.4*10=200*x=>x=2cm.

initial length=40+2=42cm

amplitude=45-42=3cm

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