    Question

# 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 simple harmonic motion of amplitude A and angular frequency ω. The total energy of the mass - spring system will be

A
12mA2ω2
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B
12mA2ω2+12ky2
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C
12ky2
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D
12mA2ω212ky2
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Solution

## The correct option is B 12mA2ω2+12ky2Let L be the relaxed length of the spring and y the extension produced in it due to force mg so that ky = mg (i) The displacement of the mass during oscillation is given by x=Asin(ωt+ϕ) At the instant when the displacement is x ​KE of mass = 12mV2=12m(dxdt)2=12mA2ω2cos2(ωt+ϕ) (iii) PE of spring = 12k(y+x)2=12k(y2+2yx+x2) =12ky2+kyx+12kx2 Using (i) and (ii) and ω=√km, we have PE of spring =12ky2+mgx+12mω2A2sin2(ωt+ϕ) (iv) taking gravitational PE at the mean position to be zero. Gravitational PE at x = -mg x (v) Adding (iii), (iv) and (v), we get Total energy of mass -spring system =12mA2ω2cos2(ωt+ϕ)+12ky2+mgx+12mω2A2sin2(ωt+ϕ)−mgx=12mA2ω2+12ky2  Suggest Corrections  0      Similar questions  Related Videos   Applying SHM
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