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Chapter 10: Problem 13

A particle executes simple harmonic motion of period \(12 \mathrm{~s}\), At \(t=0\) the particle was at right extreme position. The velocity of the particle at \(t=1 \mathrm{~s}\) is \(-2 \mathrm{~m} / \mathrm{s}\), Find the amplitude of oscillation. (a) \(\frac{24}{\pi} \mathrm{m}\) (b) \(\frac{12}{\pi} \mathrm{m}\) (c) \(\frac{9}{\pi} m\) (d) \(\frac{36}{\pi} \mathrm{m}\)

Short Answer

Expert verified
The amplitude of oscillation is \( \frac{24}{\pi} \mathrm{m} \) (Option a).

Step by step solution

01

Understand Simple Harmonic Motion (SHM) equation

The general equation for the displacement in SHM is given by \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. Since the particle starts at the right extreme position at \( t=0 \), \( x(0) = A \) which implies \( \phi = 0 \). Therefore the equation simplifies to \( x(t) = A \cos(\omega t) \).
02

Calculate Angular Frequency

The angular frequency \( \omega \) is related to the period \( T \) by the equation \( \omega = \frac{2\pi}{T} \). Given that the period \( T = 12 \) s, we have \( \omega = \frac{2\pi}{12} = \frac{\pi}{6} \mathrm{~rad/s}.\)
03

Write the Velocity Equation

The velocity of a particle in SHM is given by the derivative of displacement, \( v(t) = -A\omega \sin(\omega t) \). Here, \( v(t) = -A\frac{\pi}{6} \sin\left(\frac{\pi}{6} t\right) \).
04

Use Given Conditions to Solve for Amplitude

We know the velocity at \( t=1 \) s is \(-2 \mathrm{~m/s} \). Substitute \( t = 1 \) into the velocity equation: \[ -2 = -A\frac{\pi}{6} \sin\left(\frac{\pi}{6}\right) \]Solve for \( A \):\[ 2 = A\frac{\pi}{6} \sin\left(\frac{\pi}{6}\right) \]\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]\[ 2 = A\frac{\pi}{6} \cdot \frac{1}{2} \]\[ A = \frac{4 \cdot 6}{\pi} = \frac{24}{\pi} \mathrm{m}\]
05

Identify the Correct Answer from Options

Compare the calculated amplitude \( A = \frac{24}{\pi} \mathrm{m} \) with the given options. The correct answer is (a) \( \frac{24}{\pi} \mathrm{m} \).

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Key Concepts

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

Amplitude of Oscillation
In simple harmonic motion (SHM), the amplitude, denoted by \( A \), represents the maximum displacement of the particle from its equilibrium position. It essentially tells us how far the particle moves away from the center or resting point of its motion. Hence, the amplitude provides significant information about the extent or range of motion the oscillating particle experiences. Consider it like the height of a swing; the furthest point it reaches on either side is its amplitude.
Understanding the amplitude is crucial as it is a defining characteristic of any simple harmonic motion. Additionally, in contexts such as this exercise where the particle starts from the extreme position (maximum displacement), determining the amplitude becomes more straightforward since it coincides with the initial position value. Here, the amplitude was calculated using the velocity condition at a specific time, ensuring it aligns with the known parameters of the motion.
Angular Frequency
Angular frequency, symbolized by \( \omega \), expresses how quickly the particle oscillates in a circular manner鈥� despite the path not being a circle, the mathematics liken the motion to one. \( \omega \) gives the rate of change of the particle's phase with time and is defined by the relation \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of the motion. This ties the motion to the concept of regular circular motion typically expressed with radians.
For our exercise, with a period \( T = 12 \) seconds, the angular frequency is calculated as \( \omega = \frac{\pi}{6} \) radians per second. Angular frequency not only helps in finding the time it takes for one complete oscillation but also is vital in both displacement and velocity calculations, making it a core component of analyzing SHM. An awareness of \( \omega \)'s definition and purpose grounds one's understanding of the oscillatory process.
Velocity in SHM
The velocity of a particle undergoing simple harmonic motion dynamic can be described as the rate of change of displacement concerning time. For SHM, this velocity formula is derived from the displacement function. Given as \( v(t) = -A\omega \sin(\omega t) \), the minus sign reveals the direction of velocity is opposite to that of displacement when moving towards equilibrium.
This velocity equation highlights that velocity is maximum when the particle is at the equilibrium position due to the sine function, which ranges from \(-1\) to \(1\). Meanwhile, the velocity is zero when the particle is at the extreme positions since \( \sin(\omega t) \) equals zero at those points. In this exercise, calculating the velocity at time \( t = 1 \) second provided the necessary detail to find the amplitude because the given velocity value could be plugged into the formula, representing a strategic usage of known quantities in SHM dynamics.

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Most popular questions from this chapter

A non-uniform wire of length \(L\) and mass \(M\) has a variable density given by \(\mu=k x\) where \(x\) is the distance from one end of the wire and \(k\) is a constant. Find the time required for a pulse gencrated at one end of the wire to travel to the other end when tension in the wirc is \(T\).

Statement-1: If force acting on a particle is given by \(F=-k x^{2}\), then the motion of the particle is periodic but not the simple harmonic oscillation. Statement-2 : For the simple harmonic oscillation the force must be proportional to the displacement of the particle from the mean position.

With regard to sound waves, mark out corrcet suatements. (a) Phase difference between incident and reflected displacement waves from fixed end is zero. (b) Phase difference between incident and reflected pressure wave from fixed end is zero. (c) Phase difference between incident and reflected pressure wave from fixed end is \(\pi\). (d) Phase difference between incident and reflected displacement wave from fixed end is \(\pi\).

A stretched string is forced to transmit transverse waves by means of an oscillator coupled to one end. The string has a diameter of \(4 \mathrm{~mm}\). The amplitude of the oscillation is \(10^{-4} \mathrm{~m}\) and the frequency is \(10 \mathrm{~Hz}\). Tension in the string is \(100 \mathrm{~N}\) and mass density of wire is \(4.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Find: (a) the equation of the waves along the string, (b) the energy per unit volume of the wave, (c) the average cnergy flow per unit time across any scction of the string, (d) power required to drive the oscillator.

\(\delta=\frac{m}{L}=\frac{32}{4} \times 10^{2}=\frac{4}{5} \times 10^{2} \mathrm{~kg} / \mathrm{m}\) Now, According to question, \(v=v_{o}=220 \mathrm{II} z\) \(\therefore \frac{v}{2 L}=220 \mathrm{~Hz} \quad[\quad L=40 \mathrm{~cm}\) \(m=3.2 \mathrm{gm}\) \(\sqrt{\frac{T}{\delta}} \times \frac{1}{2 L}=220 \mathrm{II} z\) 1. \(\sqrt{\frac{T}{\rho}}=220 \times 2 L\) \(\therefore \quad \sqrt{T}=\sqrt{\rho} \times 220 \times 2 L$$$ \begin{aligned} &\therefore \quad T=\rho \times(220 \times 2 L)^{2}=\frac{4}{5} \times 10^{-2}(22 \times 2 \times 4)^{2} \\ &=\frac{4}{5} \times 10^{-2} \times 30976=2478.8 \times 10^{-2} \\ &\text { Now suress }=\frac{F}{\Lambda}=\frac{T}{\Lambda}=\frac{2478.8 \times 10^{-2}}{\left(10^{-3}\right)^{2}} \\ &=2478.8 \times 10^{4}=\mathrm{N} / \mathrm{m}^{2} \end{aligned} $$ and strain $$ =\frac{0.05}{40} \times \frac{10^{2}}{10^{2}}=\frac{5}{40} \times 10^{-2}=\frac{1}{8} \times 10^{-2} $$ \)\therefore$ (Youngs modulus) $$ \begin{aligned} Y &=\frac{\text { stress }}{\text { strain }}=\frac{2478.8 \times 10^{4}}{10^{2} / 8} \\ &=1.98 \times 101^{\prime} \mathrm{N} / \mathrm{m}^{2} \end{aligned} $$
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