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An object on the end of a spring oscillates with a frequency of \(15 \mathrm{~Hz}\). Calculate (a) the period of the motion and (b) the number of oscillations that the object undergoes in \(2 \mathrm{~min}\). SSM

Short Answer

Expert verified
The period of the motion is 0.067 s and the total number of oscillations in 2 minutes is 1800.

Step by step solution

01

Calculate the Period

To calculate the period of the motion, we can use the equation for the period, \(T = 1/f\), which is the inverse of the frequency. The frequency of the oscillations is given as 15 Hz; so, by substitifying \(f = 15 Hz\) into the equation will result the period of the oscillations \(T = 1/15 = 0.067 s\).
02

Calculate Total Oscillations

Number of oscillations over a period of time can be calculated by multiplying the frequency and the time period. Here the time period is given as 2 minutes, which is 120 seconds (since 1 minute = 60 seconds). So by substituting \(f = 15 Hz\) and \(t = 120s\) into the equation \(N=f*t\), will result the total oscillations \(N = 15Hz * 120s = 1800\).

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Key Concepts

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

Oscillation Frequency
When we talk about oscillation frequency, we're essentially discussing how often an object executes one cycle of motion within a unit of time.
In the problem at hand, the object on the spring oscillates with a frequency of 15 Hz.
In simpler terms, this means that the object completes 15 cycles or oscillations in one second.
Understanding frequency is crucial because it helps us quantify how fast a repetitive motion is happening. In the world of harmonic motion, this repetitive cycle can be observed in a variety of systems like pendulums, springs, and even electronic circuits.
The frequency is typically measured in Hertz (Hz), where 1 Hz is equivalent to one cycle per second.
  • The higher the frequency, the more cycles in a shorter period of time.
  • The frequency is directly related to how often an object oscillates back and forth.
Period of Motion
The period of motion is intimately linked to frequency. It is the time taken for one complete cycle of oscillation.
This makes it essentially the reciprocal of frequency.
In simple harmonic motion, if an object has a high frequency, it will have a short period, since it completes its oscillations rapidly. On the flip side, a lower frequency indicates a longer period.Using the formula for the period, which is given by \(T = \frac{1}{f}\), where \(T\) is the period and \(f\) is the frequency, we know that greater frequencies mean shorter periods.
  • For this specific problem, the frequency is 15 Hz, so the period \(T\) can be calculated as \(T = \frac{1}{15} \approx 0.067\) seconds.
  • Each cycle of motion is very quick, taking only 0.067 seconds.
Understanding the period gives us critical insights into the rhythm of motion cycles and when they occur.
Simple Harmonic Oscillator
A simple harmonic oscillator is a system where the restoring force acting on it is directly proportional to the displacement and acts in the opposite direction.
This is a physics model that describes many real-world systems smoothly and predictably. The spring and mass setup in this problem is a classic example of a simple harmonic oscillator.
The object moving up and down or back and forth in a regular pattern is what characterizes this type of oscillation. Here are the key properties of simple harmonic oscillators:
  • It continuously converts potential energy to kinetic energy and vice versa.
  • The motion it describes is sinusoidal in nature, leading to the repetitive cycle of movement.
  • Its frequency and period are constants, assuming no external forces like friction interfere.
Understanding simple harmonic oscillators is pivotal because they form the foundation for analyzing different phenomena in physics and engineering, such as vibrations in mechanical structures and even sound waves.

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Most popular questions from this chapter

A \( 100-\mathrm{g}\) object is fixed to the end of a spring that has a spring constant of \(15 \mathrm{~N} / \mathrm{m}\). The object is displaced \(15 \mathrm{~cm}\) to the right and released from rest at \(t=0\) to slide on a horizontal, frictionless table. (a) Find the first three times when the object is at the equilibrium position. (b) Find the first three times when the object is \(10 \mathrm{~cm}\) to the left of equilibrium. (c) What is the first time that the object is \(5 \mathrm{~cm}\) to the right of equilibrium, moving toward the left?

The period of a simple harmonic oscillator is \(0.0125 \mathrm{~s}\). What is the frequency?

An object-spring system undergoes simple harmonic motion. If the amplitude increases but the mass of the object is not changed, the total energy of the system A. increases. B. decreases. C. doesn't change. D. undergoes a sinusoidal change. E. decreases exponentially.

High tide occurs at 8:00 A.M. and is \(1 \mathrm{~m}\) above sea level. Six hours later, low tide is \(1 \mathrm{~m}\) below sea level. After another \(6 \mathrm{~h}\), high tide occurs (again \(1 \mathrm{~m}\) above sea level), then finally one last low tide ( \(6 \mathrm{~h}\) later, \(1 \mathrm{~m}\) below sea level). (a) Write a mathematical expression that would predict the level of the ocean at this beach at any time of day. (b) Find the times in the day when the ocean level is exactly at sea level.

A simple pendulum oscillates between \(\pm 8^{\circ}\) (as measured from the vertical). The length of the pendulum is \(0.50 \mathrm{~m}\). Compare the time intervals between \(\pm 8^{\circ}\) and \(\pm 4^{\circ} . \mathrm{SSM}\)

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