question_answer

The potential energy function for a particle executing simple harmonic motion is given by\[V(x)=\frac{1}{2}k{{x}^{2}}\], , where k is the force constant of the oscillator. For\[k=\frac{1}{2}N{{m}^{-1}}\] , show that a particle of total energy 1 joule moving under this potential must turn back when it reaches \[x=\pm 2m\].

Answer:

                As is known, the particle will turn back when whole of its energy is converted into potential energy, i.e.,                \[V(x)=\frac{1}{2}k{{x}^{2}}=1\left( joule \right)\] \[\frac{1}{2}\times \frac{1}{2}{{x}^{2}}=1\]        or            \[{{x}^{2}}=4\] \[x=\pm 2m\]