Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The frequency of the particle is \(\ldots \ldots \mathrm{s}^{-1}\). (A) \((1 / \pi)\) (B) \(\pi\) (C) \((1 / 2 \pi)\) (D) \((\pi / 2)\)

Short Answer

Expert verified
The frequency of the particle is \(\boxed{(A) \, \frac{1}{\pi} \, \text{s}^{-1}}\).

Step by step solution

01

Rewrite the displacement equation using the cosine identity

Let us use the identity \(\cos A \cos B + \sin A \sin B = \cos(A - B)\). Let \(A=2t\) and \(B=\alpha\). We want to find \(\alpha\) such that \(3\cos 2t + 4\sin 2t = R\cos(2t-\alpha)\), where \(R\geq 0\) The angle \(\alpha\) is given by: \(\tan \alpha = \frac{\text{coefficient of }\sin2t}{\text{coefficient of }\cos2t}\), which in our case means \(\tan \alpha = \frac{4}{3}\). Now we can find R: \(R = \sqrt{(3\cos 2t)^2 + (4\sin 2t)^2} = \sqrt{9\cos^2 2t + 16\sin^2 2t} = 5\) Thus, the displacement equation can be rewritten as: \(y = 5\cos(2t - \alpha)\)
02

Compare the simplified equation to the general harmonic equation

Now, let's compare our simplified equation to the general harmonic equation \(y = A \cos(2 \pi ft + \phi)\): \(5\cos(2t - \alpha) \, = A \cos(2 \pi ft + \phi)\)
03

Identify the frequency

Since our equation is \(y = 5\cos(2t - \alpha)\), the coefficient of \(t\) is 2. In the general harmonic equation, this is equal to \(2\pi f\): \(2 = 2 \pi f\) Solving for \(f\), we get: \(f = \frac{1}{\pi}\) Looking at the provided options, the frequency of the particle is \(\boxed{(A) \, \frac{1}{\pi} \, \text{s}^{-1}}\).

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 Vaia!

Key Concepts

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

Frequency
Frequency refers to the number of complete cycles or oscillations that a wave undergoes in a specific amount of time, typically a second. It is a key characteristic of oscillating systems, such as waves and vibrations, and is measured in Hertz (Hz), or cycles per second. In the context of our exercise, frequency determines how fast the harmonic motion is occurring. To calculate frequency from the given displacement equation, we identify the term with time variable, which in this case is multiplied by 2. In mathematical terms, this corresponds to the equation structure \(y = A \cos(2 \pi ft + \phi)\). By comparing the equations, we understand that the frequency \(f\) is directly linked to the coefficient of \(t\). Knowing that 2 is equal to \(2 \pi f\), we solve for \(f\), yielding \(f = \frac{1}{\pi}\) Hz.
Displacement Equation
The displacement equation describes how the position of a particle changes over time in harmonic motion. It typically involves trigonometric functions like sine and cosine, which relate to periodic movements such as waves.In this exercise, the displacement equation is presented as \(y = 3 \cos 2t + 4 \sin 2t\). This equation builds on the understanding that both the cosine and sine functions can describe oscillatory motion. Here, the \(3\) and \(4\) are amplitudes for the cosine and sine components, showing how far the particle moves from its mean position.By transforming these traditional trigonometric components into a simplified form \(y = R \cos(2t - \alpha)\), we make it easier to analyze by consolidating both harmonics into a single term. This transformation relies on finding the resultant amplitude \(R\) and phase \(\alpha\), which then closely resembles the format of the standard harmonic equation.
Trigonometric Identity
Trigonometric identities are fundamental tools in understanding wave equations and their transformations. They help simplify complex trigonometric expressions by using known equalities between different functions.In this exercise, we utilize the identity: \(\cos A \cos B + \sin A \sin B = \cos(A - B)\). Applying this allows us to rewrite a sum of sine and cosine terms as a single cosine or sine term with an adjusted angle. Specifically, the displacement equation \(y = 3 \cos 2t + 4 \sin 2t\) is rewritten into \(y = R \cos(2t - \alpha)\) based on this identity.This simplification is essential to convert the given equation into a form that matches the general harmonic equation, facilitating the determination of key elements like frequency and phase shift.
Harmonic Equation
The harmonic equation is a standard representation of oscillating systems, typically taking the form \(y = A \cos(2 \pi ft + \phi)\). Here, \(A\) is the amplitude, \(f\) the frequency, and \(\phi\) the phase angle.In this exercise, we compare the simplified displacement equation \(y = 5 \cos(2t - \alpha)\) to the harmonic equation. This comparison reveals important characteristics:
  • The amplitude \(A\) is 5, showing the maximum displacement.
  • The term \(2t\) suggests a frequency format where \(2 = 2 \pi f\), leading to \(f = \frac{1}{\pi}\).
  • The phase shift \(\alpha\) determines the initial positional offset in its cycle.
Understanding this format allows for a straightforward interpretation of the particle's motion, making elements like frequency easy to extract from the equation directly.

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 rectangular block having mass \(\mathrm{m}\) and cross sectional area A is floating in a liquid having density \(\rho\). If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time \(\mathrm{T}\). Then in this case \(\ldots \ldots\) (A) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{m}})\) (B) \(T \propto \sqrt{\rho}\) (C) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{A}})\) (D) \(\mathrm{T} \propto(1 / \sqrt{\rho})\)

Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is \(\ldots \ldots\) (A) 132 (B) 138 (C) 276 (D) 144

For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a) Statement \(-1\) is true, statement \(-2\) is true; statement \(-2\) is the correct explanation of statement \(-1\). (b) Statement \(-1\) is true, statement \(-2\) is true but statement \(-2\) is not the correct explanation of statement \(-1\) (c) Statement \(-1\) is true, statement \(-2\) is false (d) Statement \(-1\) is false, statement \(-2\) is true (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) d Statement \(-1:\) For a particle executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and frequency \(30 \mathrm{hz}\), the maximum acceleration is \(36 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}\). Statement \(-2:\) The maximum acceleration for the above particle is \(\pm \omega 2 \mathrm{~A}\), where \(\mathrm{A}\) is amplitude. (A) a (B) \(\mathrm{b}\) (C) \(c\) (D) \(\mathrm{d}\)

Equation for a harmonic progressive wave is given by \(\mathrm{y}=\mathrm{A}\) \(\sin (15 \pi t+10 \pi x+\pi / 3)\) where \(x\) is in meter and \(t\) is in seconds. This wave is \(\ldots \ldots\) (A) Travelling along the positive \(\mathrm{x}\) direction with a speed of \(1.5 \mathrm{~ms}^{-1}\) (B) Travelling along the negative \(\mathrm{x}\) direction with a speed of \(1.5 \mathrm{~ms}^{-1} .\) (C) Has a wavelength of \(1.5 \mathrm{~m}\) along the \(-\mathrm{x}\) direction. (D) Has a wavelength of \(1.5 \mathrm{~m}\) along the positive \(\mathrm{x}\) - direction.

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

See all solutions

Recommended explanations on English 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.

Sign-up for free