1. Inform you about time table of exam.
2. Inform you about new question papers.
3. New video tutorials information.
The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m–1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m
A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2where a = 5 m1/2 s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
- Physics Textbook Part - 1 for Class - 11 - 11086