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Chapter 15: Problem 38

A particle starts SHM from the mean position. Its amplitude is \(a\) and total energy \(E\). At one instant its kinetic energy is \(3 E / 4\) its displacement at this instant is $$ \text { (a) } y=a / \sqrt{2} $$ (b) \(y=\frac{a}{2}\) $$ \begin{array}{ll} \text { (c) } y=\frac{a}{\sqrt{3 / 2}} & \text { (d) } y=a \end{array} $$

Short Answer

Expert verified
The displacement is \(y=\frac{a}{\sqrt{2}}\), option (a).

Step by step solution

01

Understand the Problem

We are given a particle undergoing Simple Harmonic Motion (SHM) starting from the mean position. The maximum displacement (amplitude) is \(a\) and the total energy of the system is \(E\). At a certain instant, the kinetic energy is \(\frac{3E}{4}\). We need to find its displacement at this instant.
02

Formula for Total Energy in SHM

The total energy in SHM is given by \(E = \frac{1}{2}k a^2\), where \(k\) is the spring constant. This total energy is the sum of kinetic energy (KE) and potential energy (PE).
03

Relate Kinetic Energy and Displacement

In SHM, kinetic energy (KE) at a displacement \(y\) is \(KE = E - PE = E - \frac{1}{2}k y^2\). Given KE = \(\frac{3E}{4}\), we can write \(\frac{3E}{4} = E - \frac{1}{2}k y^2\).
04

Simplify the Equation

Rearrange the equation: \(\frac{1}{2}k y^2 = E - \frac{3E}{4} = \frac{E}{4}\).
05

Express in Terms of Amplitude

Given \(E = \frac{1}{2}k a^2\), substitute into the equation to get \(\frac{1}{2}k y^2 = \frac{1}{4} \cdot \frac{1}{2}k a^2\).
06

Solve for Displacement \(y\)

Cancelling out \(k\) gives \(y^2 = \frac{a^2}{2}\). Taking the square root, we find \(y = \frac{a}{\sqrt{2}}\).
07

Choose the Correct Answer

The displacement \(y = \frac{a}{\sqrt{2}}\) matches option (a). Hence, the displacement at this instant is \(\frac{a}{\sqrt{2}}\).

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Key Concepts

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

Kinetic Energy in SHM
In Simple Harmonic Motion (SHM), kinetic energy is a crucial concept. As the oscillating object moves, its speed changes, and so does its kinetic energy. The kinetic energy (KE) of a particle in SHM at a displacement of \[ y \] depends on its velocity at that point. The formula for kinetic energy in SHM is:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the particle, and \( v \) is the velocity. More conveniently, if relating it to its potential energy (PE) and total energy (E), we have:\[ KE = E - \frac{1}{2} k y^2 \] Here, \( y \) is the displacement from equilibrium, and \( k \) is the spring constant. In situations like the original exercise, where KE is known (for instance, \( \frac{3E}{4} \)), one can find the displacement \( y \).- A high kinetic energy suggests the particle is near its maximum speed, typically when \( y \) is small (around the mean position).- When KE decreases, the particle is moving towards maximum displacement, where PE is highest.
Potential Energy in SHM
Potential energy in SHM plays an equally important role as kinetic energy. At any point during the oscillation, the potential energy (PE) is stored as the system is displaced from its equilibrium position. The potential energy in SHM is calculated using the formula:\[ PE = \frac{1}{2} k y^2 \] where:- \( k \) is the spring constant.- \( y \) is the displacement from the mean position.When the particle is at the extreme positions of displacement, the entire energy of the system is potential energy. At these points, the velocity is zero, meaning kinetic energy is also zero. - As the particle returns towards the equilibrium point, the potential energy decreases, converting back into kinetic energy.- The conversion between potential and kinetic energy is what sustains the SHM.In the context of the original problem, determining the potential and kinetic energy at a given displacement can inform about the particle's position and behavior during oscillation.
Total Energy in Oscillations
In Simple Harmonic Motion, understanding the total energy is key to comprehending the system's dynamics. The total energy (\( E \)) of a system undergoing SHM remains constant if no external forces perform work on it. This energy is a combination of kinetic energy and potential energy:\[ E = KE + PE = \frac{1}{2} k a^2 \]where \( a \) is the amplitude of the motion. Here:- At the equilibrium position, total energy is all kinetic with \( KE = E \) and \( PE = 0 \).- At maximum displacement, \( PE = E \) and \( KE = 0 \).The conservation of energy in SHM signifies that the total energy \( E \) remains unchanged as the form of energy oscillates between kinetic and potential. This is important for predicting various states of the system at any time.

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Most popular questions from this chapter

An object suspended from a spring exhibits oscillations of period \(T\). Now, the spring is cut in two halves and the same object is suspended with two halves as shown in figure. The new time period of oscillation will become (a) \(\frac{T}{2 \sqrt{2}}\) (b) \(\frac{T}{2}\) (c) \(\frac{T}{\sqrt{2}}\) (d) \(2 T\)

A bottle weighing \(220 \mathrm{~g}\) and area of cross-section \(50 \mathrm{~cm}^{2}\) and height \(4 \mathrm{~cm}\) oscillates on the surface of water in vertical position. Its frequency of oscillation is (a) \(1.5 \mathrm{~Hz}\) (b) \(2.5 \mathrm{~Hz}\) (c) \(3.5 \mathrm{~Hz}\) (d) \(4.5 \mathrm{~Hz}\)

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) \(30^{\circ}\)

The bob of a simple pendulum is of mass \(10 \mathrm{~g} .\) It is suspended with a thread of \(1 \mathrm{~m} .\) If we hold the bob so as to stretch the string horizontally and release it, what will be the tension at the lowest position? \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) Zero (b) \(0.1 \mathrm{~N}\) (c) \(0.3 \mathrm{~N}\) (d) \(1.0 \mathrm{~N}\)

Let \(T_{1}\) and \(T_{2}\) be the time period of spring \(A\) and \(B\) when mass \(M\) is suspended from one end of each spring. If both springs are taken in series and the same mass \(M\) is suspended from the series combination, the time period is \(T\), then (a) \(T=T_{1}+T_{2}\) (b) \(\frac{1}{T}=\frac{1}{T_{1}}+\frac{1}{T_{2}}\) (c) \(T^{2}=T_{1}^{2}+T_{2}^{2}\) (d) \(\frac{1}{T^{2}}=\frac{1}{T_{1}^{2}}+\frac{1}{T_{2}^{2}}\)
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