Question

6. The potential energy of a simple harmonic oscillation when the article is halfway to its end point is
a) 2/3 E
b) 1/8 E
c) 1/4 E
d) 1/2 E

Answer 1

Answer:

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Potential energy of a simple harmonic oscillator

$u = \frac{1}{2} m {ω}^{2} {y}^{2}$

Kinetic energy of a simple harmonic oscillator

$k = \frac{1}{2} m {ω}^{2}( {a}^{2} - {y}^{2} )$

Here y= displacement from mean position

A= maximum displacement (or amplitude) from mean position

Total energy is

E = U + K

$= \frac{1}{2} m {ω}^{2} {y}^{2} + \frac{1}{2} m {ω}^{2} ( {a}^{2} - {y}^{2} )$

$= \frac{1}{2} m {ω}^{2} {a}^{2}$

When the particle is half way to its end point ie, at half of its amplitude then

$y = \frac{a}{2}$

Hence, potential energy

$u = \frac{1}{2} m {ω}^{2} ( \frac{a}{2} ) {}^{2}$

$= \frac{1}{4} ( \frac{1}{2} m {ω}^{2} {a}^{2} )$

u = E/4

Explanation:

Hope this may help you.....

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