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Chapter 13: Problem 28

The position of an object connected to a spring varies with time according to the expression \(x=\) \((5.2 \mathrm{~cm}) \sin (8.0 \pi t)\). Find (a) the period of this motion, (b) the frequency of the motion, (c) the amplitude of the motion, and (d) the first time after \(t=0\) that the object reaches the position \(x=2.6 \mathrm{~cm}\).

Short Answer

Expert verified
The period of the motion is \(0.25\, s\), the frequency is \(4\, Hz\), the amplitude is \(5.2\, cm\), and the first time to reach the position of \(2.6\, cm\) after \(t=0\) is \(0.125\, s\).

Step by step solution

01

Identify the Parameters

From the equation \(x = 5.2 \, \mathrm{cm} \, \sin(8.0 \pi t)\), we can see that the amplitude \(A = 5.2\, \mathrm{cm}\) and the angular frequency \(\omega = 8.0 \pi\).
02

Calculate the Period and Frequency

The period can be calculated by the formula \(T = 2\pi/\omega\). Substituting \(\omega = 8.0 \pi\) into the equation gives \(T = 2\pi/8.0\pi = 0.25\, s\). Based on this, the frequency \(f = 1/T = 4\, Hz\).
03

Determine the Amplitude

From step 1, the amplitude \(A\) was given as \(5.2 \, cm\).
04

Determine the First Time to Reach the Specific Position

To find time \(t\) when \(x = 2.6 \, cm\), we set up the equation \(2.6\, \mathrm{cm} = 5.2\, \mathrm{cm} \, \sin(8\pi t)\). Now we solve it for \(t\). First divide both sides by the amplitude to get \(\sin(8\pi t) = 0.5\). The sine function equals 0.5 at \(t = 0.125\, s\) and \(t = 0.375\, s\) within one period. But as we search for the first time \(t\) after \(t=0\), the correct answer is \(t = 0.125\, s\).

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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, the amplitude is an important concept to understand. It refers to the maximum displacement from the rest position, or equilibrium. This is essentially the furthest point the object moves during its oscillation.
In our problem, the amplitude is given right in the equation as the coefficient in front of the sine function. This means that our object swings 5.2 cm from its resting position.
  • The amplitude houses crucial information about the extent or size of the motion.
  • In our example, this value is independent of time and remains constant during the motion, assuming no external forces such as friction.
Since amplitude reflects how far the object moves from its midpoint, knowing it helps predict how large each swing or oscillation will be.
Frequency
Frequency is a key term in simple harmonic motion, describing how often the oscillations occur. It is the number of complete cycles or oscillations that happen per unit of time, generally expressed in Hertz (Hz), which means cycles per second.
To find frequency in our example, we used the formula: \[f = \frac{1}{T}\]Where \(T\) is the period of the motion. We found that \(T = 0.25 \, s\), therefore, the frequency \(f = 4 \, Hz\). This tells us that the object completes 4 full oscillations every second.
  • High frequency means more oscillations in a given time, while low frequency means fewer.
  • Frequency depends on the system, like the spring mass system in our problem.
Thus, understanding frequency helps us grasp the rhythm or the pace of the object's motion.
Period
The period in simple harmonic motion tells us the time it takes for one complete oscillation cycle. It is the duration taken by the object to return to the same point in its motion path.
In our problem, we calculated the period using the formula:\[T = \frac{2\pi}{\omega}\]where \( \omega \) is the angular frequency. From the given equation, we know \( \omega = 8.0 \pi \), leading to \( T = 0.25 \, s \). This means that every 0.25 seconds, the object completes one full cycle of motion.
The concept of period is linked closely to that of frequency since:\[f = \frac{1}{T} \quad \text{and} \quad T = \frac{1}{f}\]Keep in mind, the period depends on factors like the strength of the spring and the mass of the object.

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Most popular questions from this chapter

An object-spring system moving with simple harmonic motion has an amplitude \(A\). (a) What is the total energy of the system in terms of \(k\) and \(A\) only? (b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only the variables for the mass \(m\), velocity \(v\), spring constant \(k\), and position \(x\). (c) Using the results of parts (a) and (b) and the conservation of energy equation, find the positions \(x\) of the object when its kinetic energy equals twice the potential energy stored in the spring. (The answer should in terms of \(A\) only.)

A \(326-g\) object is attached to a spring and executes simple harmonic motion with a period of \(0.250 \mathrm{~s}\). If the total energy of the system is \(5.83 \mathrm{~J}\), find (a) the maximum speed of the object, (b) the force constant of the spring, and (c) the amplitude of the motion.

A \(2.65-\mathrm{kg}\) power line running between two towers has a length of \(38.0 \mathrm{~m}\) and is under a tension of \(12.5 \mathrm{~N}\). (a) What is the speed of a transverse pulse set up on the line? (b) If the tension in the line was unknown, describe a procedure a worker on the ground might use to estimate the tension.

A "seconds" pendulum is one that moves through its equilibrium position once each second. (The period of the pendulum is \(2.000 \mathrm{~s}\).) The length of a seconds pendulum is \(0.9927 \mathrm{~m}\) at Tokyo and \(0.9942 \mathrm{~m}\) at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

A spring oriented vertically is attached to a hard horizontal surface as in Figure P18.2. The spring has a force constant of \(1.46 \mathrm{kN} / \mathrm{m}\). How much is the spring compressed when a object of mass \(m=2.30 \mathrm{~kg}\) is placed on top of the spring and the system is at rest?
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