Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 χ= 0 to χ=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 Hz, we can find the period using the formula: T=1f=14.00=0.25 seconds
02

Identify Relevant Positions and Phases

In simple harmonic motion, the displacement χ goes from 0 to the amplitude A and back in half a period. The given positions are between χ=0 and χ=1.80 cm, which indicates a motion from the center to extreme negative displacement. This corresponds to moving from phase 0 to π/2.
03

Calculate the Time Taken

The motion from χ=0 to χ=1.80 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: Time taken=T4=0.254=0.0625 seconds

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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=1T
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=1f
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 π2. 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is 0.60 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg m2. The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length L of the clapper rod for the bell to ring silentlythat is, for the period of oscillation for the bell to equal that of the clapper?

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 kg, and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM with an amplitude of 40.0 cm and a frequency of 0.600 cycle/s. (a) The sack falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the sack instead falls off when the beam has its maximum speed, repeat part (a).

Two pendulums have the same dimensions (length L) and total mass ( m ). Pendulum A is a very small ball swinging at the end of a uniform massless bar. In pendulum B, half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

A 10.0-kg mass is traveling to the right with a speed of 2.00 m/s on a smooth horizontal surface when it collides with and sticks to a second 10.0-kg mass that is initially at rest but is attached to a light spring with force constant 170.0 N/m. (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

A 175-g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m. At the instant you make measurements on the glider, it is moving at 0.815 m/s and is 3.00 cm from its equilibrium point. Use energy conservation to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free