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3 years ago

5

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the

total mechanical energy is kinetic energy?

1 answer:

KonstantinChe [14]3 years ago

3 0

Answer:

The ratio is KE : TM = 0.75

Explanation:

from the question we are told that

The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical energy of the mass is mathematically represented as

$TM = \frac{1}{2} * k * A^2$

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

$TM = KE + PE$

=> $KE = TM - PE$

Here the potential energy of the mass is mathematically represented as

$PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

$x = \frac{A}{2}$

So

$PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

So

$KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=> $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=> $KE = 0.375 * k * A^2$

So the ratio of $KE : TM$ is mathematically represented as

$\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=> $\frac{KE}{TM} = 0.75$

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Read 2 more answers

An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k

skelet666 [1.2K]

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ( $U_{e}$ ), in joules, is equal to maximum translational kinetic energy ( $K$ ), in joules:

$U_{e} = K$

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$k$ - Spring constant, in newtons per meter.

$A$ - Amplitude, in meters.

$m$ - Object mass, in kilograms.

$v$ - Speed of the object at equilibrium, in meters per second.

If we know that $k = 450\,\frac{N}{m}$ , $m = 0.25\,kg$ and $v = 0.3\,\frac{m}{s}$ , then the amplitude of the motion is:

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k\cdot A^{2} = m\cdot v^{2}$

$A = v\cdot \sqrt{\frac{m}{k} }$

$A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }$

$A \approx 0.274\,m$

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ( $E$ ), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ( $k = 450\,\frac{N}{m}$ , $A \approx 0.274\,m$ )

$E = U_{e}$

$E = \frac{1}{2}\cdot k\cdot A^{2}$

$E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}$

$E = 16.892\,J$

The total energy of the object at any point of its motion is 16.892 joules.

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