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Chapter 14: Problem 2

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced \(0.120 \mathrm{~m}\) from its equilibrium position and released with zero initial speed, then after \(0.800 \mathrm{~s}\) its displacement is found to be \(0.120 \mathrm{~m}\) 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.

Short Answer

Expert verified
The amplititude is \(0.120 \, m\), the period is \(1.600 \, s\), and the frequency is \(0.625 \, Hz\).

Step by step solution

01

Find the Amplitude

Amplitude is the maximum displacement from the equilibrium position. From the question, it is known that the maximum displacement is \(0.120 \, m\). Therefore, the amplitude is \(0.120 \, m\).
02

Find the Period

The period (T) is the time taken for one complete cycle of vibration to pass a given point. As the mass passes the equilibrium position one time in 0.800 seconds, so that is only half of a full oscillation. Therefore, the period of the oscillation is \(0.800 \, s \times 2 = 1.600 \, s\).
03

Find the Frequency

The frequency is the number of periods or cycles per unit time and is the reciprocal of the period. That is, \(f = 1/T\). Substituting \(T = 1.600 \,s\), we find that the frequency \(f = 1/1.600 \, s^{-1} = 0.625 \, Hz\).

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Key Concepts

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

Amplitude
Amplitude, in the context of simple harmonic motion (SHM), refers to the maximum extent of displacement of an object from its central or equilibrium position. In more pictorial terms, if you picture a swing at a playground, the amplitude would be how far the swing moves from the center when it鈥檚 pushed to the farthest point on either side.

For a body oscillating on a horizontal, frictionless surface attached to a spring, like in our exercise, the amplitude is given at half the distance covered during one complete oscillation, as the motion goes from one extreme to the other. In this case, the amplitude is the given displacement of 0.120 meters (m), which means this is the farthest the object travels from its equilibrium position.

When students visualize amplitude, it is beneficial to imagine the object is moving in a straight line, where the central or equilibrium position is the starting point, and amplitude is the distance it travels away from that start before reversing direction.
Period of Oscillation
The period of oscillation, often represented by the symbol T, is a critical concept in SHM. It refers to the time it takes for an object to complete one full cycle of its motion and return to its starting point. For example, if you tap the surface of water and create ripples, the time it would take for a single ripple to expand and retreat back to the source can be likened to the period.

In the exercise, we determine that the time given (0.800 seconds) is for a half cycle, as the object passes the equilibrium position only once. A full cycle involves a complete to-and-from motion from one point of maximum amplitude, back to the center, and to the maximum amplitude on the other side. Therefore, the period is twice the given time鈥�1.600 seconds. Understanding the period helps in predicting how long a single oscillation takes, which is particularly useful in time-dependent applications, like in pendulum clocks.
Frequency
Frequency, noted as f, is intimately related to the period of oscillation and is a measure of how often a repeating event, such as a complete oscillation, occurs in a unit of time. Typically, frequency is measured in hertz (Hz), which is the number of cycles per second. If you observe a heart monitor, the frequency of the heartbeat can be seen as the number of beats per minute.

With reference to the solution, we have that the frequency is the reciprocal of the period (f = 1/T). This means that you take the number one and divide it by the period to obtain the frequency. In our case, with a period of 1.600 seconds, the frequency is 0.625 Hz. Essentially, this is saying that the object completes 0.625 cycles each second. For students, wrapping one's head around frequency is easier when thought of in terms of how many times a pendulum swings back and forth in one second.

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Most popular questions from this chapter

A block of mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant \(k\), and the amplitude of the \(\mathrm{SHM}\) is \(A\). The block has \(v=0,\) and \(x=+A\) at \(t=0 .\) It first reaches \(x=0\) when \(t=T / 4\) where \(T\) is the period of the motion. (a) In terms of \(T,\) what is the time \(t\) when the block first reaches \(x=A / 2 ?\) (b) The block has its maximum speed when \(t=T / 4\). What is the value of \(t\) when the speed of the block first reaches the value \(v_{\max } / 2 ?\) (c) Does \(v=v_{\max } / 2\) when \(x=A / 2 ?\)

DATA You hang various masses \(m\) from the end of a vertical, \(0.250 \mathrm{~kg}\) spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T=2 \pi \sqrt{m / k}\) with \(m+m_{\text {eff }},\) where \(m_{\text {eff }}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass \(m\) and measure the time for 10 complete oscillations, obtaining these data: $$ \begin{array}{l|lcccc} \boldsymbol{m}(\mathbf{k g}) & 0.100 & 0.200 & 0.300 & 0.400 & 0.500 \\ \hline \text { Time (s) } & 8.7 & 10.5 & 12.2 & 13.9 & 15.1 \end{array} $$ (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_{\text {eff }}\) of the spring's mass? (e) If a \(0.450 \mathrm{~kg}\) mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

A \(40.0 \mathrm{~N}\) force stretches a vertical spring \(0.250 \mathrm{~m}\). (a) What mass must be suspended from the spring so that the system will oscillate with a period of \(1.00 \mathrm{~s} ?\) (b) If the amplitude of the motion is \(0.050 \mathrm{~m}\) and the period is that specified in part (a), where is the object and in what direction is it moving \(0.35 \mathrm{~s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is \(0.030 \mathrm{~m}\) below the equilibrium position, moving upward?

Four passengers with combined mass \(250 \mathrm{~kg}\) compress the springs of a car with worn-out shock absorbers by \(4.00 \mathrm{~cm}\) when they get in. Model the car and passengers as a single object on a single ideal spring. If the loaded car has a period of vibration of \(1.92 \mathrm{~s}\), what is the period of vibration of the empty car?

A \(2.00 \mathrm{~kg}\) frictionless block attached to an ideal spring with force constant \(315 \mathrm{~N} / \mathrm{m}\) is undergoing simple harmonic motion. When the block has displacement \(+0.200 \mathrm{~m},\) it is moving in the negative \(x\) direction with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\). Find (a) the amplitude of the motion; (b) the block's maximum acceleration; and (c) the maximum force the spring exerts on the block.
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