Question

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 $0.5N{m}^{-1}$ . The amplitude of the particle when its total energy is 1 J is equal to

• A. $2\sqrt{2}m$
• B. 2m
• C. $\sqrt{2}m$
• D. none of the above

Answer 1

Answer: (B)

: 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 $a=2m$ .