Question

# A particle with mass 2.61 kg oscillates horizontally at the end of a horizontal spring. A...

A particle with mass 2.61 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 65 cycles of oscillation. Find the frequency, ?, the speed at the equilibrium position, ?max, the spring constant, ?, the potential energy at an endpoint, ?max, the potential energy when the particle is located 68.5% of the amplitude away from the equiliibrium position, ?, and the kinetic energy, ?, and the speed, ?, at the same position.

-A mass is hung from a vertical spring and allowed to come to rest or its equilibrium position. The mass is then pulled down an additional 2.0 m and released. As the mass oscillates, it completes one full cycle in 3.0 s . Using the given values, correctly identify the mass's amplitude, full range of vertical motion, frequency, and period. The full range of vertical motion is the distance between the maximum and minimum heights of the mass.

-The potential energy of an object attached to a spring is 2.10 J at a location where the kinetic energy is 1.30 J. If the amplitude ? of the simple harmonic motion is 22.0 cm, calculate the spring constant ? and the magnitude of the largest force ?spring, max that the object experiences.

-A 185 g object is attached to a spring that has a force constant of 72.5 N/m. The object is pulled 9.75 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table. Calculate the maximum speed of the object. maximum speed: m/s Find the locations of the object when its velocity is one-third of the maximum speed. Treat the equilibrium position as zero, positions to the right as positive, and positions to the left as negative.

-Engineers can determine properties of a structure that is modeled as a damped spring oscillator, such as a bridge, by applying a driving force to it. A weakly damped spring oscillator of mass 0.299 kg is driven by a sinusoidal force at the oscillator's resonance frequency of 35.7 Hz. Find the value of the spring constant. spring constant: N/m The amplitude of the driving force is 0.559 N and the amplitude of the oscillator's steady‑state motion in response to this driving force is 0.825 m. What is the oscillator's damping constant? damping constant:

-A uniform thin rod of length 0.825 m is hung from a horizontal nail passing through a small hole in the rod located 0.038 m from the rod's end. When the rod is set swinging about the nail at small amplitude, what is the period of oscillation? period of oscillation:

-A block is attached to a ceiling by a spring of force constant ?. When pulled down and released, the block undergoes simple harmonic motion. The motion of the block is shown in the time plot.

-A simple pendulum with a period ?=1.00 s is attached to the ceiling of an elevator which is initially at rest. Calculate the period of oscillations when the elevator begins to accelerate upwards with acceleration ?=1.50 m/s2 . period of oscillations: Calculate the period of oscillations when the elevator accelerates downwards with acceleration ?=2.00 m/s2 . period of oscillations:

-Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.52 m and exactly 7 cycles are completed in 22.2 s. When this motion is viewed as a projection of circular motion, what is the radius, r, and angular velocity, ω, of the circular motion?

(1)

Given that,

mass, m = 2.61 kg

Amplitude, A = 0.923 m

(a)

Frequency, f = 65 / 129

f = 0.5038 Hz

(b)

f = (1 / 2*pi) * sqrt (k / m)

0.5038 = (1 / 2*pi) * sqrt (k / 2.61)

k = 26.16 N/m

From energy conservation,

(1/2)kA^2 = (1/2)mv^2

speed at the equilibrium position,

vmax = sqrt (kA^2 / m) = sqrt (26.16*0.9232 / 2.61)

vmax = 2.92 m/s

(c)

Spring constant,

k = 26.16 N/m

(d)

potential energy at an endpoint,

Umax = (1/2)kA^2

Umax = (1/2)*26.16*(0.923)2

Umax = 11.14 J

(e)

Potential energy when the particle is located 68.5% of the amplitude,

U = (1/2)kx^2 = (1/2)*26.16*(0.685*0.923)2

U = 5.22 J

(f)

Kinetic energy at that position,

KE = Umax - U = 11.14 - 5.22

KE = 5.91 J

(g)

(1/2)mv^2 = 5.91

(1/2)*2.61*v^2 = 5.91

speed, v = 2.128 m/s

#### Earn Coins

Coins can be redeemed for fabulous gifts.

##### Need Online Homework Help?

Most questions answered within 1 hours.

##### Active Questions
• FBC(Pty) manufactures one product, Product X. You were given the following information for period 4. Product...
• Briefly describe the sensory and/or motor symptoms you might experience after an injury to each of...
• In contrast to excitatory postsynaptic potentials, inhibitory postsynaptic potentials… a. Involve changes in ion flow across...
• A wave on a string has a displacement according to the equation: y(x,t) = 25.0 cm...
• You are the president of high performing division and the Chief Executive Officer tells you in...