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The potential energy function for a particle executing simple harmonic motion is given by `V(x)=(1)/(2)kx^(2)`, where k is the force constant of the oscillatore. For `k=(1)/(2)Nm^(-1)`, show that a particle of total energy 1 joule moving under this potential must turn back when it reaches `x=+-2m.` |
Answer» At any instant, the total energy of an oscillator is the sum of K.E. and P.E. i.e. `E=K.E.+P.E.=(1)/(2)m u^(2)+(1)/(2)kx^(2)` The particle turns back at the instant, when its velocity becomes zero i.e. `u=0` . `:. E=0+(1)/(2)kx^(2)` As `E=1` joule and `k=(1)/(2)N//m` `:. 1=(1)/(2)xx(1)/(2)x^(2)` or `x^(2)=4, x=+-2m` |
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