IIT JEE , AIEEE Forum - Physics , Mechanics

Ank999

A spring of negligible mass having a force constant k extends by an amount y when a mass m is hung from it. The mass is pulled down a little and released. The system begins to execute SHM of amplitude A and angular frequency . The total energy of the mass-spring system will be

A) 1/2 mA² ²

B) 1/2 mA²

² + 1/2 ky²

C) 1/2 ky²

D) 1/2 mA²

² - 1/2 ky²

Ans ? (B)

Plz explain

tango_goat

in this case the equilibrium point of the mass-spring system is the intial position before the spring is strched
so total amount of the system =initial energy + energy of oscillations
=1/2 ky2 + 1/2 mA2omega2

iitkgp_bipin

At equilibrium position the spring is stretched by an amount y.
Hence at this point energy stored is (1/2)ky².

Now we introduce SHM whose amplitude is A and angular frequency is w.
Energy due to oscillations = (1/2)mA²w².

so, total energy = initial energy stored + energy due to oscillations

= (1/2)ky² + (1/2)mA²w².

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