Question

The kinetic energy and potential energy of a particle executing shm with amplitude a will be equal then its displacement is

Answer 1

Answer:

Let the displacement be x

KE=  

2

1

​

K(A  

2

−x  

2

)

PE=  

2

1

​

Kx  

2

 

Given:

KE=PE

2

1

​

K(A  

2

−x  

2

)=  

2

1

​

Kx  

2

 

x=  

2

​

 

A

​

Explanation:

Answer 2

Answer:

The kinetic energy and potential energy of a particle executing SHM with amplitude a will be equal then its displacement is $x=a/\sqrt{2}$

Explanation:

The simple harmonic motion is the motion for which the restoring force is directly proportional to the displacement.

The kinetic energy of a particle that executes simple harmonic motion is

$KE=\frac{1}{2}m\omega ^2(A^2-x^2)$

The potential energy of that particle is $PE=\frac{1}{2}m\omega ^2x^2$

where $m$ is the mass of the particle

$A$- the amplitude of the motion

$\omega ^$- angular frequency

$x$-displacement

from the question, if the kinetic energy is equal to the potential energy

$\frac{1}{2}m\omega ^2(A^2-x^2)=\frac{1}{2}m\omega ^2x^2$

$A^{2}=2x^{2} \\x=a/\sqrt{2}$ so this is the displacement of the particle