/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 If the displacement \((x)\) and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 6

If the displacement \((x)\) and velocity \((v)\) of a particle executing SHM are related through the expression \(4 v^{2}=25-x^{2}\), then its time period is (A) \(\pi\) (B) \(2 \pi\) (C) \(4 \pi\) (D) \(6 \pi\)

Short Answer

Expert verified
The time period of the particle executing simple harmonic motion with the given relationship between displacement and velocity is \(4\pi\), which corresponds to option (C).

Step by step solution

01

Rewrite the given equation in terms of velocity and displacement

The given relationship between displacement \((x)\) and velocity \((v)\) is: $$ 4v^2 = 25 - x^2 $$ Now, recall that for SHM, the following relationship holds: $$ v = \pm \omega \sqrt{A^2 - x^2} $$ Where \(A\) is the amplitude of the motion and \(\omega\) is the angular frequency. We want to find a relationship involving \(\omega\), so let's square both sides of the equation above: $$ v^2 = \omega^2(A^2 - x^2) $$
02

Compare the two equations involving \(v^2\) and find the value of \(\omega^2\)

We have these two equations involving \(v^2\): $$ 4v^2 = 25 - x^2 \\ v^2 = \omega^2(A^2 - x^2) $$ By comparing the two equations, we have: $$ 4\omega^2(A^2 - x^2) = 25 - x^2 $$ We can recognize that the amplitudes are equal if \(A^2=25\), so \(A=5\). Then the equation simplifies as follows: $$ 4\omega^2(25 - x^2) = 25 - x^2 $$ Divide both sides by \((25-x^2),\) we get: $$ 4\omega^2 = 1 $$ So, \(\omega^2 = \frac{1}{4}\).
03

Find the time period using the relationship between angular frequency and time period

We know the relationship between angular frequency \((\omega)\) and time period \((T)\) is: $$ \omega = \frac{2\pi}{T} $$ We found that \(\omega^2 = \frac{1}{4}\). So, we have $$ \left(\frac{2\pi}{T}\right)^2 = \frac{1}{4} $$ Solving for \(T\), we get: $$ T^2 = \left(\frac{4\pi}{1}\right)^2 $$ Taking the square root of both sides, we find that: $$ T = 4\pi $$ Therefore, the time period of this SHM is \(4\pi\), which corresponds to option (C).

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 91影视!

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

Two SHMs are represented by the equations: \(Y_{1}=10 \sin [3 \pi t+\pi / 4]\) \(Y_{2}=5 \cos \pi t\) (A) The amplitude ratio of the two SHM is \(1: 1\). (B) The amplitude ratio of the two SHM is \(2: 1\). (C) Time periods of both the SHMs are equal. (D) Time periods of two SHMs are different.

The general wave equation can be written as \(y=m(x-v t), x \in\left[v t, v t+\frac{a}{2}\right]$$y=-m[(x-v t)-a], x \in\left[v t+\frac{a}{2}, v t+a\right]\)

A particle of mass \(m\) is attached to a spring (of spring constant \(k\) ) and has a natural angular frequency \(\omega_{0}\). An external force \(F(t)\) proportional to \(\cos \omega t\left(\omega \neq \omega_{0}\right)\) is applied to the oscillator. The time displacement of the oscillator will be proportional to \(\quad\) (A) \(\frac{1}{m\left(\omega_{0}^{2}+\omega^{2}\right)}\) (B) \(\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\) (C) \(\frac{m}{\omega_{0}^{2}-\omega^{2}}\) (D) \(\frac{m}{\left(\omega_{0}^{2}+\omega^{2}\right)}\)

A particle of mass \(m\) moves in a straight line. If \(v\) is the velocity at a distance \(x\) from a fixed point on the line and \(v^{2}=a-b x^{2}\), where \(a\) and \(b\) are constant, then (A) The motion continues along the positive \(x\)-direction only. (B) The motion is simple harmonic. (C) The particle oscillates with a frequency equal to \(\frac{\sqrt{b}}{2 \pi}\) (D) The total energy of the particle is \(m a\).

The equation of a particle executing SHM is given by \(x=3 \cos \left(\frac{\pi}{2}\right) t \mathrm{~cm}\), where \(t\) is in second. The distance travelled by the particle in the first \(8.5 \mathrm{~s}\) is \(\left(24+\frac{n}{\sqrt{2}}\right)\) then the value of \(n\) is.
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Fluids

Read Explanation

Translational Dynamics

Read Explanation

Solid State Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Work Energy and Power

Read Explanation
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.