Question 6.4:
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.
Solution:
Total energy of the particle, E = 1 J
Force
constant, k
= 0.5 N m–1Total energy of the particle, E = 1 J
Kinetic energy of the particle, K = (1/2)mv2
According to the conservation law:
E = V + K
1 = (1/2)kx2 + (1/2)mv2
At the moment of ‘turn back’, velocity (and hence K) becomes zero.
∴ 1 = (1/2)kx2
(1/2) X 0.5x2 = 1
x2 = 4
x = ±2
Hence, the particle turns back when it reaches x = ± 2 m.
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