Question

A particle performs shm of amplitude a along a straight line.When it is at a distance of √3/2afrom mean position, it's kinetic energy gets increased by an amount 1/2mw2a2 due to an impulsive force.The. It's new amplitude becomea

Answer 1

Answer: The new amplitude will become $a' = \sqrt2a$.

Explanation:

Given that,

Distance $x = \dfrac{\sqrt3}{2}$

Increased kinetic energy $K.E = \dfrac{1}{2}m\omega^{2}a^{2}$

We know that,

The total energy is

E = K.E + P.E

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

$E = \dfrac{1}{2}m\omega^{2}a^{2}$

When particle is at a distance of $\dfrac{\sqrt 3}{2}a$ from mean position, its kinetic energy gets increased by an amount $\dfrac{1}{2}m\omega^{2}a^{2}$ due to an impulsive force.

So, the new total energy is

$E' = E + \dfrac{1}{2}m\omega^{2}a^{2}$

$\dfrac{1}{2}m\omega^{2}a'^{2} = \dfrac{1}{2}m\omega^{2}a^{2}+\dfrac{1}{2}m\omega^{2}a^{2}$

$a' = \sqrt2a$

Hence, the new amplitude will become $a' = \sqrt2a$

Answer 2

The new amplitude will become .

Explanation:

Given that,

Distance  

Increased kinetic energy  

We know that,

The total energy is

E = K.E + P.E

When particle is at a distance of  from mean position, its kinetic energy gets increased by an amount  due to an impulsive force.

So, the new total energy is

Hence, the new amplitude will become