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.

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Solution

Given that the force constant of the particle in simple harmonic motion is k=0.5 N/m and the total energy of the particle is 1 J.

The formula to calculate the potential energy of the particle is,

E= 1 2 k x m 2 x m 2 = 2E k x m = 2E k

Here, the maximum distance travelled by the particle is x m .

Substitute the values in the above equation.

x m = 2( 1 ) ( 0.5 ) = 4 =±2 m

Thus, a particle of total energy of 1 J moving under the potential, must turn back when it reaches x=±2 m.

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Q. The potential energy function for a particle executing linear simple harmonic motion is given by

U (_{x}) = \frac{1}{2} k x^{2}

, where

k

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versus

x

is shown in figure. Show that a particle of total energy

1 J

moving under this potential

^{'} t u r n s b a c k^{'}

when it reaches

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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.