Question

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm. When it is at x = x1, its kinetic energy is K =5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = -1/2 xm, the kinetic and potential energies are:

Answer 1

Answer:

8J

Explanation:

Emech=5j+3j=8j

Answer 2

Given:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm. When it is at x = 1m, its kinetic energy is K =5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J.

To find:

KE and PE when x = ½ metres?

Calculation:

$KE = \dfrac{1}{2} m { \omega}^{2} ( {a}^{2} - {x}^{2} )$

• For x = 1 , we can say:

$\implies \: 5 = \dfrac{1}{2} m { \omega}^{2} ( {a}^{2} - {1}^{2} )$

Again, PE is :

$PE = \dfrac{1}{2} m { \omega}^{2} {x}^{2}$

• For x = 1 , we can say:

$\implies 3 = \dfrac{1}{2} m { \omega}^{2} {1}^{2}$

$\implies 3 = \dfrac{1}{2} m { \omega}^{2}$

$\implies \dfrac{1}{2} m { \omega}^{2} = 3$

Putting this value in the KE Equation:

$\implies \: 5 = \dfrac{1}{2} m { \omega}^{2} ( {a}^{2} - {1}^{2} )$

$\implies \: 5 = 3( {a}^{2} -1)$

$\implies \: {a}^{2} = \dfrac{8}{3}$

Now, When x = ½ metres:

$PE = \dfrac{1}{2} m { \omega}^{2} {x}^{2}$

$\implies PE = 3 \times {( \dfrac{1}{2} )}^{2}$

$\boxed{ \implies PE = \dfrac{3}{4} \: joule}$

Now , KE when x = ½ metres:

$KE = \dfrac{1}{2} m { \omega}^{2} ( {a}^{2} - {x}^{2} )$

$\implies KE =3 \times ( \dfrac{8}{3} - \dfrac{1}{4} )$

$\implies KE =3 \times ( \dfrac{32 - 3}{12} )$

$\implies KE =3 \times ( \dfrac{29}{12} )$

$\boxed{ \implies KE =\dfrac{29}{4} \: joule}$

Hope It Helps.