Question

A particle is oscillating in SHM. What fraction of total energy is kinetic when the particle is at
the mean position (A is the amplitude of oscillation)​

Answer 1

ANSWER

In case of SHM

KE=

2

1

mω

2

(a

2

−x

2

)

and at x=a/2

KE=

2

1

mω

2

[a

2

−(a/2)

2

]

=

4

3

[

2

1

mω

2

a

2

]

Total energy =

2

1

mω

2

a

2

∴ Fraction of total energy

=

Totalenergy

KE

=

2

1

mωa

2

4

3

[

2

1

mω

2

a

2

]

=

4

3

Answer 2

Answer:

The ratio of Total Energy,E to Kinetic Energy, KE of a particle executing SHM with amplitude A at mean position is 1 .

Explanation:

• A particle is said to possess SHM if it move to and fro from a fixed point. This fixed point is called the main position of SHM.
• A particle executing SHM with amplitude A and angular velocity ω, then
• Kinetic energy of particle is given by

$Kinetic \: energy, K = \frac{1}{2} m {w}^{2} ( {A}^{2} - {y}^{2})$

• Potential energy of particle is given by

$Potential \: energy, U = \frac{1}{2} m {w}^{2} {y}^{2}$

• Total energy of particle is given by

$Total \: energy, E = KE + PE \\ = \frac{1}{2} m {w}^{2} {A}^{2}$

Given that:

• Particle executing SHM with amplitude A

To find :

• Ratio of kinetic energy to total energy at mean position i.e. y = 0

Solution:

• At mean position, Kinetic energy is given by

$Kinetic \: Energy = \frac{1}{2} m {w}^{2} {A}^{2}$

• At mean position, Kinetic energy is maximum and is equal to Total energy, E of particle.

$\frac{KE}{E} = \frac{ \frac{1}{2}m {w}^{2} {A}^{2} }{\frac{1}{2}m {w}^{2} {A}^{2}} = 1$

• Ratio of Kinetic Energy, KE to Total Energy, E is 1 .