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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 is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 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
(a) Amplitude is 0.120 m; (b) Period is 1.600 s; (c) Frequency is 0.625 Hz.

Step by step solution

01

Understand the Scenario

We have an object attached to a spring on a horizontal frictionless surface. It is displaced by 0.120 m from the equilibrium and then released with zero speed. This causes the object to oscillate back and forth.
02

Identify the Amplitude

The amplitude of oscillation is the maximum displacement from the equilibrium position. Since the object is initially displaced 0.120 m from the equilibrium, the amplitude is 0.120 m.
03

Calculate the Period of Oscillation

The time taken for the object to move from one maximum displacement to the opposite maximum displacement and back to its original position on a frictionless surface is one full cycle. Given it moves to the opposite side in 0.800 s and has passed the equilibrium once, we know half the period is 0.800 s. Thus, the full period \( T \) is twice this time: \( T = 2 \times 0.800 \text{s} = 1.600 \text{s} \).
04

Determine the Frequency

Frequency \( f \) is the reciprocal of the period \( T \). Using the calculated period: \[ f = \frac{1}{T} = \frac{1}{1.600 \text{s}} = 0.625 \text{Hz} \].

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Key Concepts

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

Oscillation Amplitude
In simple harmonic motion, the **amplitude** is a crucial concept. It defines how far an object moves from its equilibrium (rest) position. Imagine it as the highest point the object can reach on either side during its oscillation cycle. In this exercise, determining the amplitude is straightforward. The object was initially moved 0.120 meters away from the starting position when it was released from rest. This distance, 0.120 meters, represents how far it gets from its central point, or equilibrium position, either to the right or left. Therefore, the oscillation amplitude is simply 0.120 meters. Remember, the amplitude tells us just how wide the oscillation swings, but not the speed or timing of the motion.
Period of Oscillation
The **period of oscillation** is the time it takes for a complete cycle of motion. For the object with the spring, one full oscillation would mean moving from the maximum displacement on one side, passing the equilibrium, reaching the maximum on the opposite side, and returning to the starting position. In this case, we observe that it takes 0.800 seconds to go from one side to the other. Since this represents just halfway through its cycle, the total period is twice this time. Thus, the period of oscillation is 1.600 seconds.
  • The period depends on factors like the spring's stiffness, not the amplitude.
  • A longer period implies slower oscillation.
  • Period is a consistent time value that does not change with the maximum distance displaced.
Frequency of Oscillation
**Frequency** is about how often oscillations happen in a specific time frame. It is the number of complete oscillations made per second. To find frequency, take the inverse of the period. For this exercise with a period of 1.600 seconds, the frequency is calculated using the formula: \[ f = \frac{1}{T} \] This means: \[ f = \frac{1}{1.600 \text{ s}} = 0.625 \text{ Hz} \] The result, 0.625 Hertz, means the object completes 0.625 cycles every second.
  • Higher frequency means more oscillations per second.
  • Frequency and period are inversely related; as one increases, the other decreases.
  • Frequency sees practical applications in many fields, like tuning musical instruments or designing electronic circuits.

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Most popular questions from this chapter

A 0.500 -kg glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

A 2.50 -kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long. You start it swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is period on the surface of Mars, where \(g=3.71 \mathrm{m} / \mathrm{s}^{2} ?\)

An object with mass 0.200 \(\mathrm{kg}\) is acted on by an elastic restoring force with force constant 10.0 \(\mathrm{N} / \mathrm{m}\) . (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-0.300 \mathrm{m}\) to \(+0.300 \mathrm{m}\) . On your graph, let \(1 \mathrm{cm}=0.05 \mathrm{J}\) vertically and \(1 \mathrm{cm}=0.05 \mathrm{m}\) horizontally. The object is set into oscillation with an initial potential energy of 0.140 \(\mathrm{J}\) and an initial kinetic energy of 0.060 \(\mathrm{J}\) . Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant \(k\) and mass \(m\) . If the damping constant has a value \(b_{1},\) the amplitude is \(A_{1}\) when the driving angular frequency equals \(\sqrt{k / m}\) . In terms of \(A_{1},\) what is the amplitude for the same driving frequency and the same driving force amplitude \(F_{\text { max }},\) if the damping constant is \((a) 3 b_{1}\) and (b) \(b_{1} / 2 ?\)
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