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Chapter 11: Problem 20

Given that a spring oscillates at a frequency of \(4.40\) cycles per second, how long will it take to make 200 oscillations?

Short Answer

Expert verified
It takes approximately 45.46 seconds to make 200 oscillations.

Step by step solution

01

Identify the Given Information

You know the frequency of the oscillating spring, which is provided as \(4.40\) cycles per second. This is the number of oscillations the spring completes in one second.
02

Understand the Target

You need to find out how long it takes to complete \(200\) oscillations. This means you are looking for the time, in seconds, required for these oscillations given the frequency.
03

Calculate the Time for One Oscillation

The time for one complete oscillation (or cycle) is the reciprocal of the frequency. Given the frequency \(f = 4.40\, \text{Hz}\), the time period \(T\) for one oscillation is:\[ T = \frac{1}{f} = \frac{1}{4.40} \text{ seconds} \approx 0.2273 \text{ seconds} \]
04

Calculate Total Time for 200 Oscillations

To find the total time for \(200\) oscillations, multiply the time for one oscillation by the number of oscillations:\[ \text{Total Time} = 200 \times T = 200 \times 0.2273 \approx 45.46 \text{ seconds} \]

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Key Concepts

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

Frequency
Frequency is a key concept in understanding harmonic motion. It refers to how often an event occurs within a specific period of time. In the context of oscillations, it represents the number of cycles or cycles per second that an oscillating object completes. This is sometimes referred to as "hertz" and is abbreviated as Hz.

The frequency of a motion is a measure of how many times the motion repeats itself in one second. For example, in our exercise, the spring oscillates with a frequency of \(4.40\) cycles per second. This means every second the spring repeats its motion 4.4 times.

Understanding frequency helps to determine other elements of harmonic motion such as time period and total time taken for a specific number of oscillations. It is a fundamental concept that links to many other properties of waves and vibrations in physics.
Oscillations
Oscillations are repeated back-and-forth movements occurring in a regular rhythm, like the swinging of a pendulum or the vibration of a guitar string. An oscillating object moves from its position to extreme points and back repeatedly.

In simple harmonic motion, an object continues to oscillate within a certain symmetry and pattern. Each complete back-and-forth motion represents one oscillation or cycle. In the case of the spring described in the exercise, each complete movement forward and back is a single oscillation.

Analyzing oscillations allows us to understand how objects move over time and apply this knowledge to various fields such as mechanical engineering, acoustics, and electronics. They are fundamental to technologies involving sound, timekeeping, and electrical circuits.
Time Period
The time period is an essential factor in harmonic motion, representing the time it takes to complete one full oscillation. This time is calculated as the reciprocal of the frequency of the oscillation.

The relationship between time period \(T\) and frequency \(f\) is straightforward:
  • \(T = \frac{1}{f}\)
From the exercise, given a frequency of \(4.40\, \text{Hz}\), we find the time period to be approximately \(0.2273\) seconds for each oscillation.

The concept of time period enables us to determine how long it takes for a certain number of oscillations to occur. Knowing the time period allows us to calculate the total time required for any number of oscillations by simply multiplying the time period with the number of oscillations. For example, this was used to find that \(200\) oscillations take around \(45.46\) seconds.

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Most popular questions from this chapter

In an old gasoline engine, a piston undergoes vertical SHM with an amplitude of \(7.0 \mathrm{~cm}\). A washer rests on top of the piston. As the motor speed is slowly increased, at what frequency will the washer no longer stay in contact with the piston? The situation we are looking for is when the maximum downward acceleration of the washer equals that of free fall, namely, \(g\). If the piston accelerates down faster than that, the washer will lose contact. In SHM, the acceleration is given in terms of the displacement and the period as $$ a=-\frac{4 \pi^{2}}{T^{2}} x $$ (To see this, notice that \(a=-\mathrm{F} / \mathrm{m}\). But from \(T=2 \pi \sqrt{\mathrm{m} / k}\), we have \(k=4 \pi^{2} m / T^{2}\), which then gives the above expression for \(a\).) With the upward direction chosen as positive, the largest downward (most negative) acceleration occurs for \(x=+x_{0}=0.070 \mathrm{~m}\); it is $$ a_{0}=\frac{4 \pi^{2}}{T^{2}}(0.070 \mathrm{~m}) $$ The washer will separate from the piston when \(a_{0}\) first becomes equal to \(g\). Therefore, the critical period for the SHM, \(T_{c}\), is given by $$ \frac{4 \pi^{2}}{T_{c}^{2}}(0.070 \mathrm{~m})=g \quad \text { or } \quad T_{c}=2 \pi \sqrt{\frac{0.070 \mathrm{~m}}{g}}=0.53 \mathrm{~s} $$ This corresponds to the frequency \(f_{c}=1 / T_{c}=1.9 \mathrm{~Hz}\). The washer will separate from the piston if the piston's frequency exceeds \(1.9\) cycles/s.

A "seconds pendulum" beats seconds; that is, it takes \(1 \mathrm{~s}\) for half a cycle. (a) What is the length of a simple "seconds pendulum" at a place where \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} ?(b)\) What is the length there of \(a\) pendulum for which \(T=1.00 \mathrm{~s}\) ?

A 200-g mass vibrates horizontally without friction at the end of a horizontal spring for which \(k=7.0 \mathrm{~N} / \mathrm{m} .\) The mass is displaced \(5.0\) \(\mathrm{cm}\) from equilibrium and released. Find \((a)\) its maximum speed and \((b)\) its speed when it is \(3.0 \mathrm{~cm}\) from equilibrium. ( \(c\) ) What is its acceleration in each of these cases? where \(k=7.0 \mathrm{~N} / \mathrm{m}, x_{0}=0.050 \mathrm{~m}\), and \(m=0.200 \mathrm{~kg} .\) Solving for \(|v|\) gives $$ |v|=\sqrt{\left(x_{0}^{2}-x^{2}\right) \frac{k}{m}} $$ (a) The speed is a maximum when \(x=0\); that is, when the mass is passing through the equilibrium position: $$ |v|=x_{0} \sqrt{\frac{k}{m}}=(0.050 \mathrm{~m}) \sqrt{\frac{7.0 \mathrm{~N} / \mathrm{m}}{0.200 \mathrm{~kg}}}=0.30 \mathrm{~m} / \mathrm{s} $$ (b) When \(x=0.030 \mathrm{~m}\), $$ |v|=\sqrt{\frac{7.0 \mathrm{~N} / \mathrm{m}}{0.200 \mathrm{~kg}}\left[(0.050)^{2}-(0.030)^{2}\right] \mathrm{m}^{2}}=0.24 \mathrm{~m} / \mathrm{s} $$ (c) Using \(F=m a\) and \(F=k x\), $$ a=\frac{k}{m} x=\left(35 \mathrm{~s}^{-2}\right)(x) $$ which yields \(a=0\) when the mass is at \(x=0\), and \(a=1.1 \mathrm{~ms}^{2}\) when \(x=0.030 \mathrm{~m}\).

A particle oscillates according to the equation \(y=5.0 \cos 23 t\), where \(y\) is in centimeters. Find its frequency of oscillation and its position at \(t=0.15 \mathrm{~s}\).

A 2.5-kg body undergoes SHM and makes exactly 3 vibrations each second. Compute the acceleration and the restoring force acting on the body when its displacement from the equilibrium position is \(5.0 \mathrm{~cm}\).
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