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What is the maximum acceleration of a platform that oscillates at amplitude \(2.20 \mathrm{~cm}\) and frequency \(6.60 \mathrm{~Hz}\) ?

Short Answer

Expert verified
The maximum acceleration is approximately 37.835 m/s².

Step by step solution

01

Identify and Convert Given Values

We are given the amplitude of oscillation as \(2.20\, \mathrm{cm}\) and the frequency as \(6.60\, \mathrm{Hz}\). First, we need to convert the amplitude to meters: \(2.20\, \mathrm{cm} = 0.0220\, \mathrm{m}\).
02

Understand the Formula for Maximum Acceleration

The maximum acceleration \(a_{max}\) of an oscillating platform can be calculated using the formula:\[a_{max} = (2\pi f)^2 A\]where \(f\) is the frequency and \(A\) is the amplitude.
03

Calculate \(2\pi f\)

Calculate the term \(2\pi f\). Since \(f = 6.60\, \mathrm{Hz}\), we find:\[2\pi f = 2\times \pi \times 6.60 \approx 41.469\]
04

Compute \((2\pi f)^2\)

Square the result from Step 3 to get:\[(2\pi f)^2 = (41.469)^2 \approx 1719.79\]
05

Calculate Maximum Acceleration

Substitute the values into the formula for \(a_{max}\):\[a_{max} = (2\pi f)^2 \times A = 1719.79 \times 0.0220 \approx 37.835\, \mathrm{m/s^2}\]
06

Conclusion

The maximum acceleration of the platform is approximately \(37.835\, \mathrm{m/s^2}\).

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Key Concepts

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

Amplitude
In oscillatory motion, the amplitude is a key parameter that defines the maximum extent of an object's displacement from its rest position. Imagine watching a swing move back and forth; the amplitude is the maximum height it reaches on either side before reversing direction.
Here's why amplitude matters:
  • It represents the energy in the system: Larger amplitude means higher energy.
  • It affects the distance traveled during oscillation.
In our given problem, the amplitude was initially given as 2.20 cm. Since oscillatory motion calculations often use meters, it was converted to 0.0220 m for ease in calculations.
Keep in mind that amplitude only tells you about the furthest position reached, not how fast or how often that position is reached. That's the role of frequency.
Frequency
Frequency is another crucial parameter in understanding oscillatory motion. It refers to how many complete cycles of motion occur in one second. The higher the frequency, the more times the oscillatory system moves back and forth in a given timeframe.
Let's explore the implications of frequency:
  • Frequency is measured in Hertz (Hz), where 1 Hz equals one cycle per second.
  • It determines the perceived pitch of a sound in acoustics and affects the rate of vibration in mechanical systems.
In our exercise, the platform has a frequency of 6.60 Hz, signifying that the platform oscillates 6.60 times each second. This plays a direct role in determining how quickly the maximum acceleration is reached during each cycle.
Frequency's profound effect is evident in how it, together with amplitude, defines the dynamic nature of oscillatory systems.
Maximum Acceleration
Maximized acceleration in an oscillating system tells us the peak rate at which the velocity changes, both as the object moves to and from the maximum displacement. It's calculated using the formula: \[a_{max} = (2\pi f)^2 A\]This captures the relationship between frequency and amplitude.Let's clarify what maximum acceleration entails:
  • Acceleration measures how fast the speed changes; maximum acceleration tells us this at the point where it's highest.
  • The formula shows maximum acceleration depends on both the oscillation frequency and amplitude.
In our example, the maximum acceleration was calculated to be approximately 37.835 m/s². This value was derived by plugging in the frequency (6.60 Hz) and the amplitude (0.0220 m) into our formula.
Maximum acceleration is an indicator of how intense the motion can get, and it's particularly useful in engineering to ensure structural safety and efficacious operation of oscillating devices.

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Most popular questions from this chapter

A grandfather clock has a pendulum that consists of a thin brass disk of radius \(r=15.00 \mathrm{~cm}\) and mass \(1.000 \mathrm{~kg}\) that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. \(15-56 .\) If the pendulum is to have a period of 2.000 s for small oscillations at a place where \(g=9.800 \mathrm{~m} / \mathrm{s}^{2},\) what must be the rod length \(L\) to the nearest tenth of a millimeter?

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