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A graph of potential energy V (x) verses x is shown in Figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

Answers (1)

The graph above is between kinetic energy and x

As we know from the law of conservation of energy,

E = KE + PE

E = KE + V(x)  

KE = E - V(x)

The potential energy is maximum at point A,

PE = E0

Hence, KE = E0 -E0 = 0

At point B,

Let PE = V_b, x = 0

KE = E _0- V_{0}

At point C, PE = 0

X = x_{1}

Hence, KE = E - V(x) =E0 - 0 =E0

At point D,

X = x_{2}, KE = E_0

At point E,

X = x_{3}, PE = E_0

Hence, KE = 0

(ii)

The graph is between velocity versus x

Here, KE = -mv

V = \sqrt{\frac{2K}{M}}

V = \sqrt{K}

Through point A and F, KE = 0

X_{a} = 0, X_{d} = x_{3}

At point C and D,

X_{c} = x_{1}, X_{d} = x_{2}

KE = E0

Hence V_{max} = +/- \sqrt{E_0} = V_{0}

At point B, x = 0 and KE = E_{1}

So, V_{b} = \sqrt{E_{1}} = +/- V

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infoexpert24

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