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1 month ago

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When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic

energy?

2 answers:

Ostrovityanka [3.2K]1 month ago

6 0

Answer:

$KE: TE = 3: 4$

Explanation:

We know that the comprehensive mechanical energy of an object performing SHM is expressed as

$E_{total} = \frac{1}{2}KA^2$

where it is established that

A = the amplitude of SHM

K = spring constant

it is understood that the total mechanical energy in the spring remains constant, hence we conclude

kinetic energy + Potential energy = total mechanical energy

the potential energy of the spring at a certain position is defined as

$U = \frac{1}{2}kx^2$

at the specified position x = A/2

$U = \frac{1}{2}K(\frac{A}{2})^2 = \frac{1}{8}KA^2$

now we arrive at

$KE + \frac{1}{8}KA^2 = \frac{1}{2}KA^2$

$KE = \frac{3}{8}KA^2$

the ratio of kinetic energy to total mechanical energy will thus be calculated as

$KE: TE = \frac{3}{8}KA^2: \frac{1}{2}KA^2$

$KE: TE = 3: 4$

ValentinkaMS [3.4K]1 month ago

5 0

Total energy associated with a spring:
$E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2$

When x = 0.5a:
$\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2$

The ratio:
$\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}$

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