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Chapter 15: Problem 4

Two pendulums have time period \(T\) and \(5 T / 4\). They start SHM at the same time from the mean position. What will be the phase difference between them after the bigger pendulum completed one oscillation? (a) \(45^{\circ}\) (b) \(90^{\circ}\) (c) \(60^{\circ}\) (d) \(30^{\circ}\)

Short Answer

Expert verified
The phase difference is \(90^{\circ}\). Option (b).

Step by step solution

01

Determine Complete Cycles for Bigger Pendulum

The bigger pendulum completes exactly one full oscillation, which by definition means it undergoes a phase change of \(360^{\circ}\) or \(2\pi\) radians.
02

Calculate Time Elapsed for Bigger Pendulum

The time taken for the bigger pendulum to complete one oscillation is its time period, which is given by \(\frac{5T}{4}\). This is the time over which both pendulums have been oscillating.
03

Calculate Oscillations of Smaller Pendulum in Same Time

During this time of \(\frac{5T}{4}\), we need to calculate how many oscillations the smaller pendulum completes. The time period of the smaller pendulum is \(T\), so it completes \(\frac{\frac{5T}{4}}{T} = \frac{5}{4}\) oscillations.
04

Convert Fractional Oscillations to Phase

The smaller pendulum completing \(\frac{5}{4}\) oscillations means it has undergone \(5/4\times360^{\circ} = 450^{\circ}\) of phase shift.
05

Calculate Phase Difference

Subtract the phase of the full oscillation (\(360^{\circ}\)) from the phase of \(450^{\circ}\) to find the phase difference: \[450^{\circ} - 360^{\circ} = 90^{\circ}.\]

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Key Concepts

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

Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of oscillatory motion where an object moves back and forth over a fixed path. The motion is repetitive and occurs in a straight line.
The key characteristics of SHM include:
  • The object has a maximum displacement from its mean position on both sides, called the amplitude.
  • The motion has a specific time period, which is the time it takes to complete one full oscillation.
  • In SHM, the restoring force is directly proportional to the displacement but in the opposite direction, given by Hooke's Law: \( F = -kx \).
The negative sign indicates that the force is always directed back toward the mean position. When studying pendulum motion or vibrations in strings or springs, SHM provides a great approach to understanding their dynamics.
Pendulum Oscillation
A pendulum undergoes an oscillation pattern defined by the gentle back and forth swinging motion often seen in clocks or swings. This motion is propelled by the force of gravity and the pendulum's support at one fixed point.
To evaluate pendulum oscillations, consider these important aspects:
  • The pendulum moves in SHM provided the angles are small. This means the motion is sinusoidal and periodic.
  • The time period of a simple pendulum, a special type, can be approximated as \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
  • The motion is initiated with the pendulum being lifted to a certain height from which it is released, causing it to swing.
Pendulums can be found in various applications, and understanding their oscillation is essential for syncing their use in mechanisms.
Time Period in Oscillations
The time period in oscillations, often denoted by \( T \), is a fundamental concept in harmonic motion. It refers to the duration required for an object to complete one full oscillation cycle.
In understanding this concept, consider the following:
  • The time period remains constant for a given oscillation under uniform conditions, making it essential in predicting future motion.
  • For simple pendulums, the time period is mostly affected by the pendulum's length and the local gravitational force.
  • In resonant systems, such as springs and pendulums, the time period's predictability allows for precise timing devices like clocks.
Knowing the time period helps in calculating the frequency of oscillations, expressed as \( f = \frac{1}{T} \). This gives the number of cycles an oscillating object completes per second, directly tied to the fundamental understanding of oscillatory systems.

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Most popular questions from this chapter

A heavy sphere of mass \(m\) is suspended by string of length \(l\). The sphere is made to revolve about a vertical line passing through the point of suspension in a horizontal circle such that the string always remains inclined to the vertical at an angle \(\theta\). What is its period of revolution? (a) \(T=2 \pi \sqrt{\frac{l}{g}}\) (b) \(T=2 \pi \sqrt{\frac{l \cos \theta}{g}}\) (c) \(T=2 \pi \sqrt{\frac{l \sin \theta}{g}}\) (d) \(T=2 \pi \sqrt{\frac{l \tan \theta}{g}}\)

Displacement-time equation of a particle executing SHM is, \(x=4 \sin \omega t+3 \sin (\omega t+\pi / 3)\). Here \(x\) is in centimetre and \(t\) in second. The amplitude of oscillation of the particle is approximately (a) \(5 \mathrm{~cm}\) (b) \(6 \mathrm{~cm}\) (c) \(7 \mathrm{~cm}\) (d) \(9 \mathrm{~cm}\)

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is \(T\). With what acceleration should the lift be accelerated upwards in order to reduce its period to T/2? ( \(g\) is the acceleration due to gravity) [Karnataka CET 2008] (a) \(4 \mathrm{~g}\) (b) \(g\) (c) \(2 \mathrm{~g}\) (d) \(3 \bar{g}\)

A particle of mass \(m\) is executing oscillations about the origin on the \(x\)-axis with amplitude \(A\). Its potential energy is given as \(U(x)=\alpha x^{4}\), where \(\alpha\) is positive constant. The \(x\)-coordinate of mass where potential energy is one-third of the kinetic energy of particle, is (a) \(\pm \frac{A}{\sqrt{3}}\) (b) \(\pm \frac{A}{\sqrt{2}}\) (c) \(\pm \frac{A}{3}\) (d) \(\pm \frac{A}{2}\)

A uniform cylinder of length \(L\) and mass \(M\) having cross-sectional area \(A\) is suspended with its vertical length, from a fixed point by a massless spring, such that it is half submerged in a liquid of density \(d\) at equilibrium position. When released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is \(k\), the frequency of oscillation of the cylinder is $$ \frac{1}{2 \pi}\left(\frac{k-A d g}{M}\right)^{1 / 2} \quad \text { (b) } \frac{1}{2 \pi}\left(\frac{k+A d g}{M}\right)^{1 / 2} $$ (c) \(\frac{1}{2 \pi}\left(\frac{k-d g L}{M}\right)^{1 / 2}\) (d) \(\frac{1}{2 \pi}\left(\frac{k+A g L}{A d g}\right)^{1 / 2}\)
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