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Two particles undergo SHM along the same line with the same time period \((T)\) and equal amplitudes \((A)\). At a particular instant, one particle is at \(x=-A\) and the other is at \(x=0\). They move in the same direction. They will cross each other at time \(t\) and at position \(x\) then (A) \(t=\frac{4 T}{3}\) (B) \(t=\frac{3 T}{8}\) (C) \(x=\frac{A}{2}\) (D) \(x=\frac{A}{\sqrt{2}}\)

Short Answer

Expert verified
The answers are (A) \(t=\frac{3 T}{8}\) and (C) \(x=\frac{A}{2}\).

Step by step solution

01

Understand the given conditions

There are two particles, both undergoing a simple harmonic motion along the same line. Both have same time period \(T\) and same amplitude \(A\). One particle is at \(x=-A\) and the other one is at \(x=0\) at a particular instant. Both are moving in the same direction.
02

Formulate the position equations

\[x_1=A\sin(\omega t+\phi_1)\] \[x_2=A\sin(\omega t+\phi_2)\] where \(x_1\) and \(x_2\) are the positions of the two particles, \(\omega\) is the angular velocity, \(t\) is the time, and \(\phi_1\), \(\phi_2\) are the initial phases. Given the initial conditions, replace \(x_1=-A\) and \(x_2=0\), solve the system to get \(\phi_1\) and \(\phi_2\).
03

Calculate the time when particles cross each other

The particles cross each other if their positions become equal, which means \(x_1(=A\sin(\omega t+\phi_1))=x_2(=A\sin(\omega t+\phi_2))\). Substitute the initial phases from step 2 into this equation and solve it to get the time \(t\) when they cross each other.
04

Calculate the position when the particles cross each other

Substitute \(t\) we get from step 3 into either of the position equations we get from step 2, which will give us the position \(x\) when the particles cross each other.

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Key Concepts

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

SHM Equations
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. The equations governing SHM are foundational for understanding the motion of particles in such a system. The position of a particle in SHM can be described by the equation:
\[x(t) = A \sin(\omega t + \phi)\]
where \(x(t)\) is the displacement at time \(t\), \(A\) is the maximum displacement from the equilibrium position known as the amplitude, \(\omega\) is the angular velocity, and \(\phi\) is the initial phase angle. This equation provides us with the displacement of the particle at any given time, which is essential when analyzing the motion of the two particles in the given problem.
Time Period
The time period of a particle in SHM, denoted by \(T\), is the amount of time it takes to complete one full oscillation or cycle. This is a crucial concept as it relates to the frequency of the motion with which the particle goes back and forth. The time period is inversely proportional to the angular frequency \(\omega\) and is given by the formula:
\[T = \frac{2\pi}{\omega}\]
For the two particles described in the exercise, knowledge of the time period is valuable because it allows us to compare the positions of the particles at various instances within their cycles, leading us to determine when and where they will cross paths.
Amplitude
Amplitude, often represented by the symbol \(A\), is the maximum extent of displacement of the particle from its equilibrium position in SHM. It determines the range within which the particle oscillates and is a measure of the energy in the motion. The amplitude is a key indicator of how far the particle will travel from the center position before reversing direction. In the context of our exercise, both particles have the same amplitude, which simplifies the analysis because their maximum displacements are equal, although their starting positions and phase angles may differ.
Phase Difference
The concept of phase difference, represented as \(\Delta \phi\), is used to describe the relative positions in the cycles of two oscillating particles. It is defined as the difference in their phase angles. Phase difference dictates whether the particles are in phase, out of phase, or somewhere in between. When particles have a phase difference of \(\pi\) radians (180 degrees), they are in antiphase, meaning when one is at maximum positive displacement, the other is at an equivalent negative displacement. This concept is central to predicting when particles will be at the same position, as their phase difference will indicate if and when their paths intersect, a situation explored in the exercise at hand.
Angular Velocity
Angular velocity, denoted by \(\omega\), is a measure of how quickly the particle is oscillating in SHM. It is calculated by the equation \(\omega = \frac{2\pi}{T}\), where \(T\) is the time period. Angular velocity gives us the rate at which the angle (in terms of radians) changes with time, revealing the speed of the oscillation. In our exercise, knowing the angular velocity is necessary to plug into our SHM equations to describe the particles' motion as they travel across their paths, ultimately helping to solve for the time \(t\) and position \(x\) at which they cross.

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Most popular questions from this chapter

The equation of a wave on a string of linear mass density \(0.04 \mathrm{~kg} \mathrm{~m}^{-1}\) is given by $$ y=0.02(m) \sin \left[2 \pi\left(\frac{t}{0.04(s)}-\frac{x}{0.50(m)}\right)\right] \text {. The ten- } $$ sion in the string is (A) \(4.0 \mathrm{~N}\) (B) \(12.5 \mathrm{~N}\) (C) \(0.5 \mathrm{~N}\) (D) \(6.25 \mathrm{~N}\)

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