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Chapter 9: Problem 9

When a particle oscillates simple harmonically, its kinetic energy varies periodically. If frequency of the particle is \(n\), the frequency of the kinetic energy is (A) \(n / 2\) (B) \(n\) (C) \(2 n\) (D) \(4 n\)

Short Answer

Expert verified
The frequency of the kinetic energy of a particle oscillating with simple harmonic motion is twice its original frequency, so the correct answer is (C) \(2n\).

Step by step solution

01

Determine velocity of particle in SHM

If a particle is oscillating harmonically, its position \(x(t)\) as a function of time follows the equation: \(x(t) = A\cos(2\pi nt + \phi)\) where \(A\) is the amplitude, \(\phi\) is the initial phase, and \(t\) is the time in seconds. Next, let's find the velocity of the particle, which can be found by taking the first derivative of the position function with respect to time: \(v(t) = \frac{dx(t)}{dt}\)
02

Differentiate position equation

We will differentiate \(x(t) = A\cos(2\pi nt + \phi)\) with respect to time: \(v(t) = -A(2\pi n)\sin(2\pi nt + \phi)\)
03

Determine kinetic energy

Recall that the kinetic energy of a particle of mass \(m\) can be determined by the equation: \(K = \frac{1}{2}mv^2\) Plugging in the velocity expression from step 2, we get: \(K = \frac{1}{2}m(-A[2\pi n]\sin(2\pi nt + \phi))^2\) Simplifying the equation gives us: \(K = 2\pi^2 mAn^2\sin^2(2\pi nt + \phi)\)
04

Analyze the frequency of kinetic energy

Now let's analyze the frequency of the kinetic energy expression. Notice that the kinetic energy depends on the sin^2 term, which can be rewritten in terms of cosines using a double-angle formula: \(\sin^2(2\pi nt + \phi) = \frac{1}{2}(1 - \cos(4\pi nt + 2\phi))\) With this form, it becomes apparent that the frequency of the kinetic energy is double that of the original frequency \(n\). The frequency of the kinetic energy is \(4\pi nt\), compared to the particle's oscillation frequency of \(2\pi nt\). So, the correct answer is: (C) \(2 n\)

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Most popular questions from this chapter

If \(Y_{1}=5 \sin (\omega t)\) and \(Y_{2}=5[\sqrt{3} \sin \omega t+\cos \omega t]\) are two SHMs, the ratio of their amplitude is (A) \(1: \sqrt{3}\) (B) \(1: 3\) (C) \(1: 2\). (D) \(1: \cos \left(\frac{\pi}{6}\right)\)

A simple pendulum is suspended from the ceiling of an empty box falling in air near earth surface. The total mass of system is \(M\). The box experiences air resistance \(\vec{R}=-k \vec{v}\), where \(v\) is the velocity of box and \(k\) is a positive constant. After some time, it is found that period of oscillation of pendulum becomes double the value when it would have suspended from a point on earth. The velocity of box at that moment \(v=\frac{M g}{n k}\), then the value of \(\mathrm{n}\) is. (Take \(g\) in air same as on earth's surface.)

Assertion: The scalar product of the displacement and the acceleration in SHM is never greater than zero. Reason: Acceleration is linearly proportional to and opposite to displacement. (A) \(\mathrm{A}\) (B) B (C) \(\bar{C}\) (D) D

Sound waves of wavelength \(\lambda\) travelling in a medium with a speed of \(v \mathrm{~m} / \mathrm{s}\) enter into another medium where its speed in \(2 v \mathrm{~m} / \mathrm{s}\). Wavelength of sound waves in the second medium is (A) \(\lambda\) (B) \(\frac{\lambda}{2}\) (C) \(2 \lambda\) (D) \(4 \lambda\)

A signal of \(5 \mathrm{kHz}\) frequency is amplitude modulated on a carrier wave of frequency \(2 \mathrm{MHz}\). The frequencies of the resulting signal is/are (A) \(2005 \mathrm{kHz}\), and \(1995 \mathrm{kHz}\) (B) \(2005 \mathrm{kHz}, 2000 \mathrm{kHz}\) and \(1995 \mathrm{kHz}\) (C) \(2000 \mathrm{kHz}\) and \(1995 \mathrm{kHz}\) (D) \(2 \mathrm{MHz}\) only
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