1. Tardigrade
  2. Question
  3. Physics
  4. The potential energy function of a particle executing linear simple harmonic motion is given by U(x)=(1/2)kx2, where k the force constant of the oscillator is equal to 0.5 text Nm-1 . The amplitude of the particle when its total energy is 1 J is equal to

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Q. The potential energy function of a particle executing linear simple harmonic motion is given by $ U(x)=\frac{1}{2}k{{x}^{2}}, $ where k the force constant of the oscillator is equal to $ 0.5\text{ }N{{m}^{-1}} $ . The amplitude of the particle when its total energy is 1 J is equal to

AMUAMU 1997

Solution:

: Total energy of particle executing SHM $ =\frac{1}{2}m{{\omega }^{2}}{{a}^{2}} $ Potential energy of particle $ =\frac{1}{2}m{{\omega }^{2}}{{x}^{2}} $ $ \frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}} $ or $ k=m{{\omega }^{2}} $ $ \therefore $ Total energy $ =\frac{1}{2}m{{\omega }^{2}}{{a}^{2}} $ Or, $ 1=\frac{k{{a}^{2}}}{2} $ or $ {{a}^{2}}=\frac{2}{k}=\frac{2}{0.5}=\frac{4}{1} $ or $ a=2\text{ }m $ .