Physics, 15.04.2020 03:48 pattyv9845

A particle with mass 2.29 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.923 m and a duration of 129 s for 74 cycles of oscillation. Find the frequency, f , the speed at the equilibrium position, v max , the spring constant, k , the potential energy at an endpoint, U max , the potential energy when the particle is located 36.1 % of the amplitude away from the equiliibrium position, U , and the kinetic energy, K , and the speed, v , at the same

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Answer from: BriTheKueen

a) The frequency of the particle is 0.574 hertz.

b) The spring constant of the particle is 29.777 newtons per meters.

c) The speed at the equilibrium position is approximately 3.328 meters per second.

d) The elastic potential energy is 12.681 joules.

e) The potential energy when the particle is located 36.1 % of the amplitude is 1.651 joules.

The kinetic energy when the particle is located 36.1 % of the amplitude is 11.03 joules.

The speed when the particle is located at 36.1 % of the amplitude is approximately 3.104 meters per second.

The particle experiments a simple harmonic motion, whose angular frequency ( $\omega$ ), in radians per second, is described by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

- Spring constant, in newtons per meter.

- Mass, in kilograms.

a) The frequency (), in hertz, is the number of cycles done in a second, that is to say:

$f = \frac{n}{t}$ (2)

Where:

- Number of cycles.

- Time associated to a number of cycles.

If we know that n = 74 and $t = 129\,s$ , then the frequency is:

$f = 0.574\,hz$

The frequency of the particle is 0.574 hertz.

b) The angular frequency is also found by the following expression:

$\omega = 2\pi\cdot f$

$\omega = 2\pi\cdot (0.574\,hz)$

$\omega \approx 3.606\,\frac{rad}{s}$

The spring constant (), in newtons per meter, is found by (1):

$k = \omega^{2}\cdot m$

$k = \left(3.606\,\frac{rad}{s} \right)^{2}\cdot (2.29\,kg)$

$k = 29.777\,\frac{N}{m}$

The spring constant of the particle is 29.777 newtons per meters.

c) The speed at the equilibrium position (), in meters per second, occur when translational kinetic energy reaches its maximum. By the principle of energy conservation, such maximum kinetic energy equals a maximum elastic potential energy:

$\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{2}\cdot m\cdot v^{2}$ (3)

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

Where is the amplitude, in meters.

If we know that $k = 29.777\,\frac{N}{m}$ , $m = 2.29\,kg$ and $A = 0.923\,m$ , then the speed at the equilibrium position is:

$v = \sqrt{\frac{29.777\,\frac{N}{m} }{2.29\,kg} }\cdot (0.923\,m)$

$v \approx 3.328\,\frac{m}{s}$

The speed at the equilibrium position is approximately 3.328 meters per second.

d) The potential energy (), in joules, at an endpoint is the maximum possible elastic potential energy:

$U = \frac{1}{2}\cdot k\cdot A^{2}$ (4)

If we know that $k = 29.777\,\frac{N}{m}$ and $A = 0.923\,m$ , then the potential energy is:

$U = \frac{1}{2}\cdot \left(29.777\,\frac{N}{m} \right)\cdot (0.923\,m)^{2}$

$U = 12.681\,J$

The elastic potential energy is 12.681 joules.

e) The potential energy when the particle is located 36.1 % of the amplitude away from the equilibrium position: ( $k = 29.777\,\frac{N}{m}$ , $A = 0.333\,m$ )

$U = \frac{1}{2}\cdot \left(29.777\,\frac{N}{m} \right)\cdot (0.333\,m)^{2}$

$U = 1.651\,J.$

The potential energy when the particle is located 36.1 % of the amplitude is 1.651 joules.

By the principle of energy conservation, we derive an expression of the translational kinetic energy:

$K = U_{max}-U$ (5)

Where:

$U_{max}$ - Maximum potential energy, in joules.

- Potential energy, in joules.

- Kinetic energy, in joules.

If we know that $U_{max} = 12.681\,J$ and $U = 1.651\,J.$ , then the kinetic energy of the particle is:

$K = 12.681\,J-1.651\,J$

$K = 11.03\,J$

The kinetic energy when the particle is located 36.1 % of the amplitude is 11.03 joules.

And the speed (), in meters per second, is described by the following formula:

$v = \sqrt{\frac{2\cdot K}{m} }$ (6)

If we know that $K = 11.03\,J$ and $m = 2.29\,kg$ , then the speed of the particle is:

$v = \sqrt{\frac{2\cdot (11.03\,J)}{2.29\,kg} }$

$v\approx 3.104\,\frac{m}{s}$

The speed when the particle is located at 36.1 % of the amplitude is approximately 3.104 meters per second.

We kindly invite to check this question on principle on energy conservation: link

Answer from: graymonky12

Energy at

Spring Constant, k = 29.75N/m

x =

at 36.1%U = 1.65 Joules

Explanation:

Given

Mass of Particle, m = 2.29kg

Amplitude, A = 0.923m

Period, T = 129/74 -sec/cycles

Number of oscillation, n = 74 cycles

Find the frequency, f ,

Frequency the rate per second of a vibration constituting a wave;

Frequency is calculated by solving for the inverse of period

i.e. $f = \frac{1}{T}$ , Substitute T = 129/74 -sec/cycles

$f =\frac{1}{\frac{129}{74}}$

$f = \frac{74}{129}$

f = 0.57364341085 Hz

f = 0.5734 Hz ----- Approximated

Calculating the spring constant, k

To solve this, we make use of period formula as follows

$T = 2 pi \sqrt{\frac{m}{k} }$

Make k the subject of formula

First, take square root of both sides

$T^{2} = \frac{4pi^{2}{m}}{k}}$

Multiply through by $\frac{k}{T^{2} }$ ; This gives

$k = \frac{4pi^{2}{m}}{T^{2}}}$

Substitute values

$k = \frac{4 * pi^{2} * 2.29 }{(\frac{129}{74})^{2} }$

k = 29.75N/m

Calculating the energy at Umax

The energy at Umax is a kinetic energy and is given as $0.5kA^{2}$

by Substitution

Energy at Umax = $0.5 * 29.85 * 0.923^{2}$

Energy at Umax = 12.71 Joules

Calculating Speed at equilibrium position

Using $K.E = 0.5mv^{2}$

Where K.E = Umax = 12.71

By Substitution

$12.71 = 0.5 * 2.29 * v^{2}$

vmax = 3.33 m/s

36.1 % of the amplitude away from the equilibrium position is

x = 0.923 * 36.1 % =

x = 0.333203m at 36.1%

At that point the elastic potential energy is calculated by U = $0.5kA^{2}$

U = $0.5 * 29.75 * 0.333203^{2}$

U = 1.65 Joules

The Kinetic Energy at this point is the difference between Energy at Umax and U

At that point the kinetic energy is:

K.E = Umax - U

Substitute Values

K.E = 12.71 - 1.65

K.E = 11.06 Joules

Lastly, the velocity

We calculate velocity from $K.E = 0.5mv^{2}$

Substitute values

$11.06 = 0.5 * 2.29 * v^{2}$

$9.79 = v^{2}$

$v = \sqrt{9.79}$

v = 3.13 m/s

Answer from: Quest

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Answer from: Quest

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A particle with mass 2.29 kg oscillates horizontally at the end of a horizontal spring. A student me...