Verified Solution • 2025–26 Session moderate

Q5.4: NCERT Physics - Part 1 WORK, ENERGY AND POWER 5.4 Solution: The potential energy function for a particle executing linear…

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx_/2, where k is the force constant of the oscillator. For k = 0.5 N m__, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = � 2 m.

Step-by-Step Verified Solution

**Answer:** At x = �2 m, V = 1 J = total energy, so KE = 0 and the particle turns back. **Step-by-Step Explanation:** Total energy E = 1 J, k = 0.5 N/m. At turning point, KE = 0, so E = V(x):
1=12kx2=12(0.5)x2=0.25x21 = \frac{1}{2}kx^2 = \frac{1}{2}(0.5)x^2 = 0.25x^2
x2=4  ⟹  x=±2 mx^2 = 4 \implies x = \pm 2 \text{ m}
Hence the particle turns back at x = �2 m.

All Questions in this Chapter

Q5.1 Q5.2 Q5.3 Q5.4 Q5.5 Q5.6 Q5.7 Q5.8 Q5.9 Q5.10 Q5.11 Q5.12 Q5.13 Q5.14 Q5.15 Q5.16 Q5.17 Q5.18 Q5.19 Q5.20 Q5.21 Q5.22 Q5.23

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