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Chapter 9: Problem 5

The ratio of kinetic energy at mean position to the potential energy when the displacement is half of the amplitude is (A) \(\frac{4}{1}\) (B) \(\frac{2}{3}\) (C) \(\frac{4}{3}\) (D) \(\frac{1}{2}\)

Short Answer

Expert verified
The ratio of kinetic energy at mean position to the potential energy when the displacement is half of the amplitude is \(\frac{4}{1}\).

Step by step solution

01

Write down the formula for Kinetic energy and Potential energy

Using the formula for kinetic energy (KE) for a simple harmonic oscillator, we can write, \(KE = \frac{1}{2}mv^2\) And the potential energy (PE) formula: \(PE = \frac{1}{2}kx^2\) Where m is the mass, v is the velocity, k is the spring constant, and x is the displacement.
02

Express velocity in terms of amplitude and angular frequency

In a simple harmonic motion, we have the relation between velocity, maximum displacement (amplitude A), and angular frequency (蠅) as: \(v = \omega \sqrt{A^2 - x^2}\)
03

Calculate kinetic energy at mean position

At mean position, the displacement (x) is equal to 0. So, using the formula found in step 2, \(v_{mean} = \omega A\) Now, substituting this velocity value in the kinetic energy formula, \(KE_{mean} = \frac{1}{2}m (\omega A)^2\)
04

Calculate potential energy when displacement is half of the amplitude

When the displacement (x) is half of the amplitude (A/2), \(PE_{A/2} = \frac{1}{2} k(\frac{A}{2})^2\) From the relation between spring constant (k), mass (m), and angular frequency (蠅), \(k = m\omega^2\) Substituting the value of k in the potential energy formula, we get: \(PE_{A/2} = \frac{1}{2} m\omega^2(\frac{A}{2})^2\)
05

Find the ratio between kinetic energy at mean position and potential energy when displacement is half of the amplitude

Now, we find the ratio: \(\frac{KE_{mean}}{PE_{A/2}} = \frac{\frac{1}{2}m(\omega A)^2}{\frac{1}{2}m\omega^2(\frac{A}{2})^2}\) Simplifying the equation, we get: \(\frac{KE_{mean}}{PE_{A/2}} = \frac{(\omega A)^2}{\omega^2(\frac{A}{2})^2} = \frac{4}{1}\) So, the ratio of kinetic energy at mean position to potential energy when the displacement is half of the amplitude is 4:1. Hence, the correct option is: (A) \(\frac{4}{1}\)

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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 track followed for two perpendicular SHMs is a perfect ellipse when \(\left(\delta\right.\)-phase difference, \(A_{1}, A_{2}\) amplitudes) (A) \(\delta=\frac{\pi}{4}, A_{1} \neq A_{2}\) (B) \(\delta=\frac{3 \pi}{4}, A_{1} \neq A_{2}\) (C) \(\delta=\frac{\pi}{2}, A_{1} \neq A_{2}\) (D) \(\delta=\pi, A_{1}=A_{2}\)

Two monochromatic coherent point sources \(S_{I}\) and \(S_{2}\) are separated by a distance \(L .\) Each source emits light of wavelength \(\lambda\), where \(L \gg \lambda\). The line \(S_{1} S_{2}\) when extended meets a screen perpendicular to it at a point \(A\). (A) The interference fringes on the screen are circular in shape. (B) The interference fringes on the screen are straight lines perpendicular to the line \(S_{1} S_{2} A\). (C) The point \(A\) is an intensity maxima if \(L=n \lambda\). (D) The point \(A\) is always an intensity maxima for any separation \(L\).

A body executes simple harmonic motion. The potential energy (PE), the kinetic energy (KE) and total energy (TE) are measured as a function of displacement \(x\). Which of the following statements is true ? (A) \(\mathrm{KE}\) is maximum when \(x=0\). (B) TE is zero when \(x=0\) (C) \(\mathrm{KE}\) is maximum when \(x\) is maximum (D) \(\mathrm{PE}\) is maximum when \(x=0\)

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) \(1: \cos \left(\frac{\pi}{6}\right)\)
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