/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 (I) A pendulum makes 32 oscillat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(I) A pendulum makes 32 oscillations in exactly 50 s. What is its \((a)\) period and \((b)\) frequency?

Short Answer

Expert verified
The period is 1.56 s; the frequency is 0.641 Hz.

Step by step solution

01

Understand the Problem

We are given that a pendulum makes 32 oscillations in 50 seconds. We need to find two things: the period and the frequency of the pendulum's oscillations.
02

Calculate the Period

The period of a pendulum is the time it takes to make one complete oscillation. We can calculate the period \( T \) by dividing the total time by the number of oscillations. Thus, \( T = \frac{50 \text{ s}}{32} \).
03

Perform the Calculation for Period

Calculating \( T = \frac{50}{32} \), we find the period \( T \approx 1.56 \text{ s} \). Thus, each oscillation takes approximately 1.56 seconds.
04

Calculate the Frequency

The frequency \( f \) of a pendulum is the number of oscillations it makes per second. It is the inverse of the period, \( f = \frac{1}{T} \).
05

Perform the Calculation for Frequency

Now, calculate \( f = \frac{1}{1.56 \text{ s}} \approx 0.641 \text{ Hz} \). This means the pendulum oscillates approximately 0.641 times per second.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Oscillation
An oscillation refers to the repetitive movement of a system between two positions over time, typically in a back-and-forth manner. Imagine a pendulum that sways to and fro; each complete cycle from one side to the other, and back again, is one oscillation.
Oscillations can occur in various systems, not just pendulums—think of the swings in a playground, heartbeats, or even sound waves. But in all cases, the core concept remains the same: a repeated motion following a specific path.
In the context of a pendulum, understanding oscillations helps to comprehend how energy is transferred and conserved over time. This oscillatory motion is crucial in many fields, such as mechanics and wave theory, because it describes the behavior of a multitude of physical systems.
The beauty of oscillations lies in their predictability and regularity, making them easy to measure and crucial to studying other dynamic systems. With pendulums, each oscillation is a consistent time unit we can use to investigate other properties, like frequency and period.
Exploring Frequency
Frequency, a fundamental concept in the study of oscillations, tells us how often these oscillations occur within a certain time frame. Specifically, frequency indicates how many complete cycles of oscillation occur in one second.
It is measured in hertz (Hz), named after the physicist Heinrich Hertz. Thus, a frequency of 1 Hz means that one complete oscillation occurs every second. For our pendulum example, a frequency of approximately 0.641 Hz signifies that the pendulum swings through its complete back-and-forth movement 0.641 times in one second.
Calculating frequency is straightforward when you know the period (the time for one oscillation). By taking the inverse of the period \( f = \frac{1}{T} \), we derive the frequency. This relationship implies that as the time per oscillation (period) decreases, the system oscillates more frequently, resulting in a higher frequency.
Understanding frequency is crucial in fields such as telecommunications and musical acoustics, where it defines signal transmission and sound pitches respectively. It provides a clear metric for describing how fast or slow an oscillating system behaves.
The Concept of Period
The period of an oscillation is the time it takes for one full cycle of movement to complete. In other words, it measures how long it takes for a pendulum, or any oscillating system, to move through its entire path and return to its starting position.
The formula to determine the period \( T \) is given by dividing the total time by the number of oscillations \( T = \frac{time}{oscillations} \). Using the example provided, a pendulum taking 50 seconds to complete 32 oscillations yields a period of approximately 1.56 seconds.
Grasping the concept of the period is essential for predicting and controlling systems in engineering and physics. Knowledge of the period lets engineers design rhythms in machinery, synchronize signals, and apply periodicity concepts in wave-based technologies.
Additionally, the period is a crucial parameter in determining a system's frequency, illustrating how these two important measurements are intrinsically connected. Together, period and frequency form a framework for understanding cycles in dynamic systems, ensuring that both time and speed are comprehensively accounted for.

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 physical pendulum consists of an \(85-\mathrm{cm}\) -long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig. 14-38). The motion is damped because of friction in The motion is damped because of friction in the pivot; the damping force is approximately proportional to \(d \theta / d t\). The rod is set in oscillation by displacing it \(15^{\circ}\) from its equilibrium position and releasing it. After \(8.0 \mathrm{~s}\) the amplitude of the oscillation has been reduced to \(5.5^{\circ} .\) If the angular displacement can be written as \(\theta=A e^{-\gamma t} \cos \omega^{\prime} t,\) find (a) \(\gamma,\) (b) the approximate period of the motion, and \((c)\) how long it takes for the amplitude to be reduced to \(\frac{1}{2}\) of its original value.

In Section 5 of "oscillations," the oscillation of a simple pendulum (Fig. 46\()\) is viewed as linear motion along the arc length \(x\) and analyzed via \(F=m a\) . Alternatively, the pendulum's movement can be regarded as rotational motion about its point of support and analyzed using \(\tau=I \alpha\) . Carry out this alternative analysis and show that \(\theta(t)=\theta_{\max } \cos \left(\sqrt{\frac{g}{\ell}} t+\phi\right)\) where \(\quad \theta(t)\) is the angular displacement of the pendulum from the vertical at time \(t,\) as long as its maximum value is less than about \(15^{\circ}\) .

(III) A glider on an air track is connected by springs to either end of the track (Fig. \(39 ) .\) Both springs have the same spring constant, \(k,\) and the glider has mass \(M .\) (a) Determine the frequency of the oscillation, assuming no damping, if \(k=125 \mathrm{N} / \mathrm{m}\) and \(M=215 \mathrm{g} .\) (b) It is observed that after 55 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. Estimate the value of \(\gamma,\) using Eq. \(16 .(c)\) How long does it take the amplitude to decrease to one- quarter of its initial value? \(x=A e^{-\gamma t} \cos \omega^{\prime} t\)

A pendulum has a period of \(1.35 \mathrm{~s}\) on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

(I) A 1.15 -kg mass oscillates according to the equation \(x=0.650 \cos 7.40 t\) where \(x\) is in meters and \(t\) in seconds. Determine \((a)\) the amplitude, \((b)\) the frequency, \((c)\) the total energy, and \((d)\) the kinetic energy and potential energy when \(x=0.260 \mathrm{m} .\)

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.