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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
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B
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C
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D
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Solution

The correct option is B
Let 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

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