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Work, Energy, and Power — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power3 Marks
Question
The potential energy function for a particle executing linear simple harmonic motion is given by $\text{V(x)} = \frac{\text{kx}^2}{2},$ where k is the force constant of the oscillator. For $k = 0.5Nm^{-1}$, the graph of V(x) versus x is shown in Show that a particle of total energy 1J moving under this potential must ‘turn back’ when it reaches $\text{x} =\pm 2\text{m}.$
✓

Answer

Given, Potential energy for a particle executing linear simple harmonic motion is, $\text{V(x)}=\frac{1}{2}\text{kx}^2$
Where, $\text{k}=\frac{1}{2}\text{N/m}$

Total energy of particle $E = 1J$
Since at extreme position total energy is potential energy,
$\therefore\text{V}=\frac{1}{2}\text{kx}^2=1\text{J}$
$\Rightarrow\text{x}=\pm\sqrt{\frac{2}{\text{k}}}=\pm\sqrt{4}$
$\Rightarrow\text{x}=\pm2\text{m}$ Hence the results.

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