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An object is undergoing SHM with period 1.200 s and amplitude 0.600 m. At t= 0 the object is at x= 0 and is moving in the negative x-direction. How far is the object from the equilibrium position when t= 0.480 s?

Short Answer

Expert verified
The object is approximately 0.571 m from the equilibrium position at t = 0.480 s.

Step by step solution

01

Formula for SHM Displacement

The equation for the displacement of an object in simple harmonic motion (SHM) is given by:x(t)=Acos(ωt+ϕ)where:- A is the amplitude.- ω is the angular frequency.- ϕ is the phase constant.Given that A=0.600 m and the period T=1.200 s, we need to find ω.
02

Calculate Angular Frequency

The angular frequency ω can be calculated using the formula:ω=2πTSubstitute T=1.200 s:ω=2π1.200=5π3 rad/s
03

Determine Phase Constant φ

At t=0, the object is at x=0 and moving in the negative x-direction, which suggests a sine function is more suitable for this condition. Therefore, we use:x(t)=Asin(ωt)However, initially using cosine, for consistency, we have to use:x(0)=Acos(ϕ)=0which implies ϕ=π2 since cos(π2)=0. Adjusting the equation indicates the object is indeed moving in the negative direction.
04

Apply SHM Equation at t = 0.480 s

Substitute t=0.480 s and the values A=0.600, ω=5π3, ϕ=π2 into the displacement equation:x(0.480)=0.600cos(5π3×0.480+π2)
05

Simplify and Calculate

Calculate the inside of the cosine:5π3×0.480=0.800πThen add the phase constant:0.800π+π2=0.800π+0.500π=1.300πNow evaluate the cosine:x(0.480)=0.600cos(1.300π)=0.600cos(π+0.300π)=0.600(cos(0.300π))Where cos(0.300π)0.9511, so,x(0.480)=0.600×0.95110.5707 m
06

Find Distance from Equilibrium

The distance from the equilibrium position is the absolute value of the displacement:|x(0.480)|=|0.5707|0.5707 m

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
In simple harmonic motion (SHM), the amplitude represents the maximum displacement of the object from its equilibrium position. In essence, it illustrates how far the object moves on either side of this central point.
In our exercise, the amplitude is given as 0.600 m. This means the object can move up to 0.600 meters away from the equilibrium before changing directions.
  • The greater the amplitude, the larger the displacement of the object from equilibrium.
  • Amplitude is crucial as it influences the energy of the system. A higher amplitude means more energy.
When solving problems involving SHM, always remember that while amplitude affects the total energy, it does not influence the period or frequency of the motion.
Angular Frequency
Angular frequency (ω) in SHM is a measure of how quickly the object moves through its cycle. It is related to the oscillation period and is expressed in radians per second.
To find angular frequency, use the equation:ω=2πTwhere T is the period of oscillation. In this exercise, with T=1.200 s, the angular frequency calculates to:ω=2π1.200=5π3rad/sUl bullet points can be really helpful here!
  • Angular frequency helps determine the speed of oscillations.
  • It's independent of the amplitude but crucial for understanding how cycles progress over time.
Thus, learn always to calculate the angular frequency to comprehend better how fast the system oscillates.
Phase Constant
The phase constant (ϕ) in the context of SHM determines the starting position of the oscillating object relative to its equilibrium. It effectively shifts the wave horizontally on a graph.
Our exercise provides insights into why determining the phase constant is essential. Since the object starts at x=0 and moves negatively, we use the cosine form,Acos(ϕ)=0resulting in ϕ=π2.Switching functions or solving for ϕ depends on initial conditions:
  • The phase constant is often chosen to match initial displacement and velocity conditions.
  • A correct phase constant aligns theoretical models to actual movements.
Understanding phase constants allow for more accurate predictions of position over time.
Cosine Function
In SHM, the cosine function is often used to model the displacement of an object over time. The function:x(t)=Acos(ωt+ϕ)captures how position changes with time using amplitude, angular frequency, and phase constant.
In this exercise, translating theoretical equations to practical values involved:
  • Substituting given quantities such as amplitude, angular frequency, and phase constant.
  • Evaluating the equation at specific times to find its current position.
Ultimately, cosine functions are pivotal for calculating how far an object is from equilibrium at any given moment, making it central to analyzing SHM.

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Most popular questions from this chapter

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k= 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s. What are the toy's (a) total energy at any point of its motion; (b) amplitude of motion; (c) maximum speed during its motion?

Two uniform solid spheres, each with mass M= 0.800 kg and radius R= 0.0800 m, are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant k= 160 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

An 85.0-kg mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 m long. He holds one end of the rope, and the other end is tied higher up on a rock face. Since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. At the lowest point of his swing, he plans to let go and drop a short distance to the ground. (a) How long after he begins his swing will the climber first reach his lowest point? (b) If he missed the first chance to drop off, how long after first beginning his swing will the climber reach his lowest point for the second time?

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