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A machine part is undergoing SHM with a frequency of 4.00 Hz and amplitude 1.80 cm. How long does it take the part to go from \(\chi =\) 0 to \(\chi = -1.80\) cm?

Short Answer

Expert verified
0.0625 seconds.

Step by step solution

01

Determine the Total Period of SHM

The period \( T \) of a simple harmonic motion is the reciprocal of the frequency \( f \). Given the frequency \( f = 4.00 \text{ Hz} \), we can find the period using the formula: \[ T = \frac{1}{f} = \frac{1}{4.00} = 0.25 \text{ seconds} \]
02

Identify Relevant Positions and Phases

In simple harmonic motion, the displacement \( \chi \) goes from \( 0 \) to the amplitude \( A \) and back in half a period. The given positions are between \( \chi = 0 \) and \( \chi = -1.80 \text{ cm} \), which indicates a motion from the center to extreme negative displacement. This corresponds to moving from phase \( 0 \) to \( -\pi/2 \).
03

Calculate the Time Taken

The motion from \( \chi = 0 \) to \( \chi = -1.80 \text{ cm} \) covers a quarter of the full cycle (as it moves from the mean position to the negative extreme, which is a quarter of oscillation). The total period is 0.25 seconds, hence the journey from \( 0 \) to the negative displacement takes: \[ \text{Time taken} = \frac{T}{4} = \frac{0.25}{4} = 0.0625 \text{ seconds} \]

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Key Concepts

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

Frequency
In simple harmonic motion (SHM), frequency tells us how quickly the system is oscillating. It is typically measured in Hertz (Hz), which is the number of complete cycles per second. For example, if the frequency is 4.00 Hz, this means the system completes 4 full cycles every second.

The frequency can be used to understand how fast the object is moving back and forth. If you have a high frequency, like 10 Hz, the object moves rapidly. Conversely, a lower frequency of 2 Hz signifies slower movement. If you know the time period of SHM, you can also find the frequency using the formula:
  • Formula: Frequency, \( f = \frac{1}{T} \)
where \( T \) is the period of the oscillation. This relationship is vital in comprehending how these properties interact with each other in harmonic motion.
Amplitude
Amplitude is a fundamental aspect of simple harmonic motion, indicating the maximum distance an object moves from its central position. Measured in meters (m) or centimeters (cm), it reflects the "size" of the motion. For instance, an amplitude of 1.80 cm means the object swings or moves 1.80 cm from its equilibrium point at its peak.

Amplitude does not affect the frequency or period of the oscillation; instead, it provides insight into the energy of the system. Larger amplitudes often suggest more energy within the system:
  • A higher amplitude in a swinging pendulum means it swings wider.
  • For a spring, a larger amplitude indicates further stretch or compression with each oscillation.
Recognizing the amplitude is vital for visualizing the reach of the oscillating object and understanding its movement extremes.
Period
The period in simple harmonic motion refers to the time it takes to complete one full cycle of motion. It is the duration from one point in the cycle back to that same point in the next cycle. In the provided exercise, the period is found using the inverse of the frequency:
  • Formula: \( T = \frac{1}{f} \)
Given that the frequency is 4.00 Hz, the period becomes 0.25 seconds.

Period helps in understanding how long each oscillation lasts. Whether it's a pendulum or a vibrating string, a smaller period means quicker oscillations. Period is a critical parameter in timing aspects of oscillations, letting us plan and calculate other phases of motion accurately.
Phase
Phase in the context of simple harmonic motion helps pinpoint the specific stage or position of the system within its cycle at a given time. It provides an understanding of the position of the moving object at any point in time, usually expressed in radians or degrees.

Think of phase as a snapshot of the oscillation's progress. In the given example, when the part moves from the center to the negative extreme, it corresponds to a phase change from 0 to \(-\frac{\pi}{2}\). This quarter-cycle shift is particularly useful for determining the timing and motion between various points in the cycle.

Phase differences can also represent the lead or lag between different oscillating systems or parts, clearing the path for complex motion analysis. Understanding phase helps in predicting future positions and syncing systems that rely on harmonic motion.

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Most popular questions from this chapter

An object is undergoing SHM with period 1.200 s and amplitude 0.600 m. At \(t =\) 0 the object is at \(x =\) 0 and is moving in the negative \(x\)-direction. How far is the object from the equilibrium position when \(t =\) 0.480 s?

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is \(x = +\)9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found by using the energy approach used in Section 14.3, and which cannot? Explain.

A 2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/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.

The wings of the blue-throated hummingbird \((Lampornis\) \(clemenciae)\), which inhabits Mexico and the southwestern United States, beat at a rate of up to 900 times per minute. Calculate (a) the period of vibration of this bird's wings, (b) the frequency of the wings' vibration, and (c) the angular frequency of the bird's wing beats.

A 50.0-g hard-boiled egg moves on the end of a spring with force constant \(k =\) 25.0 N/m. Its initial displacement is 0.300 m. A damping force \(F_x = -bv_x\) acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant \(b\).

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