Question

The total vibrational energy of a particle in S.H.M.is E. Its kinetic energy at half the amplitude frommean position will be :(1) E/2 (2) E/3 (3) E/4 (4) 3E/4​

Answer 1

Explanation:

1-E/2 is the answer to the question.

Answer 2

Its kinetic energy at half the amplitude from mean position will be 3E/4.

Explanation:

The kinetic energy of the vibration is given by :

$E=\dfrac{1}{2}kA^2$

Kinetic energy at a particular position is given by :

$E=\dfrac{1}{2}k(A^2-x^2)$

When x = A/2

New kinetic energy,

$E'=\dfrac{1}{2}k(A^2-(\dfrac{A}{2})^2)$

$E'=\dfrac{1}{2}k(A^2-\dfrac{A^2}{4})$

$E'=\dfrac{3}{4}(\dfrac{1}{2}kA^2)$

$E'=\dfrac{3}{4}E$

So, Its kinetic energy at half the amplitude from mean position will be 3E/4. Hence, this is the required solution.

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Simple harmonic motion

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