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

Rajasthan 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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