/*! 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 59 The total energy of a simple har... [FREE SOLUTION] | 91Ó°ÊÓ

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The total energy of a simple harmonic oscillator is proportional to [Kerala CET 2008] (a) square root of displacement (b) velocity (c) frequency (d) amplitude (e) square of the amplitude

Short Answer

Expert verified
(e) square of the amplitude

Step by step solution

01

Understand the Formula for Energy in Simple Harmonic Motion

The total energy of a simple harmonic oscillator in SHM (simple harmonic motion) is given by the formula: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( E \) is the total energy, \( m \) is the mass, \( \omega \) is the angular frequency, and \( A \) is the amplitude.
02

Identify Proportional Components

From the formula, notice that the amplitude \( A \) is squared, indicating that energy is directly proportional to the square of the amplitude. This means \( E \propto A^2 \).
03

Review the Options

The problem provides options: (a) square root of displacement, (b) velocity, (c) frequency, (d) amplitude, and (e) square of the amplitude. Based on the derived energy formula, the energy is proportional to the square of the amplitude.
04

Choose the Correct Answer

Comparing all the options, the correct answer since energy is proportional to \( A^2 \) is option (e) square of the amplitude.

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Key Concepts

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

Total Energy of Harmonic Oscillator
The total energy of a harmonic oscillator in simple harmonic motion (SHM) can be thought of as the sum of its potential and kinetic energies. This energy is conserved, meaning that it does not change over time, allowing the oscillator to continue its periodic motion. The formula that represents the total energy in SHM is:
  • \[ E = \frac{1}{2} m \omega^2 A^2 \]
Here, \( E \) represents the total energy, \( m \) denotes the mass of the oscillator, \( \omega \) is the angular frequency, and \( A \) signifies the amplitude of the motion. The elegance of SHM lies in its predictable nature, where each variable plays a crucial role in determining the motion's characteristics.
The total energy remains constant even though the distribution between potential and kinetic energy might vary as the oscillator moves. This constancy gives rise to smooth oscillations without friction or external energy influences.
Proportionality in SHM
Proportionality plays a vital role in understanding how different aspects of motion in SHM are interrelated. In the context of the total energy formula for a harmonic oscillator, we see that energy is heavily influenced by particular parameters—a standout being the relationship with the amplitude.
  • The energy is directly proportional to the square of the amplitude (\( A^2 \)).
This means that as the amplitude increases, the total energy of the oscillator increases quadratically, not linearly. For instance, doubling the amplitude results in a fourfold increase in energy. This proportionality highlights how critical amplitude is in affecting the oscillator's total energy.
Moreover, understanding these relationships can help in predicting and controlling the behavior of oscillators in practical applications. It's fascinating to note how such a simple algebraic expression can translate into real-world phenomena.
Energy Formula in SHM
The energy formula in SHM gives us a deep insight into how energy is distributed and conserved in a simple harmonic oscillator. To break it down:
  • The formula is \[ E = \frac{1}{2} m \omega^2 A^2 \]
  • It consists of kinetic energy, which is highest during zero displacements, and potential energy, which peaks at maximum displacement.
Each component of the formula serves a purpose:
  • Mass \( m \) affects inertia and how quickly the oscillator can respond to forces.
  • Angular frequency \( \omega \) relates to how fast the oscillator cycles through its motion.
  • Amplitude \( A \) measures the extent of motion, having the greatest impact on energy since it is squared.
This equation provides a complete snapshot of the oscillator's motion dynamics. There's a beauty in how this seemingly simple formula interconnects all elements of oscillation within the realm of classical physics. Understanding this can be foundational in fields ranging from engineering to natural sciences where oscillatory motion is key.

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Most popular questions from this chapter

A point mass is subjected to two simultaneous sinusoidal \(\quad\) displacement in \(\quad X\)-direction \(X_{1}(t)=A \sin \omega t\) and \(X_{2}(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right) .\) Adding a third sinusoidal displacement \(X_{3}(t)=B \sin (\omega t+\phi)\) brings the mass to a complete rest. The value of \(B\) and \(\phi\) (a) \(\sqrt{2} A, \frac{3 \pi}{4}\) (b) \(A, \frac{4 \pi}{3}\) (c) \(\sqrt{3} A, \frac{5 \pi}{6}\) (d) \(A, \frac{\pi}{3}\)

Two pendulums of length \(121 \mathrm{~cm}\) and \(100 \mathrm{~cm}\) start vibrating. At same instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in phase at the mean position? (a) 10 (b) 11 (c) 20 (d) 21

A coin is placed on a horizontal platform, which undergoes horizontal SHM about a mean position \(O\). The coin placed on platform does not slip, coefficient of friction between the coin and the platform is \(\mu\). The amplitude of oscillation is gradually increased. The coil will begin to slip on the platform for the first time (a) at the mean position (b) at the extreme position of oscillations (c) for an amplitude of \(\mu g / \omega^{2}\) (d) for an amplitude of \(g / \mu \alpha^{3}\)

Two points are located at a distance of \(10 \mathrm{~m}\) and \(15 \mathrm{~m}\) from the source of oscillation. The period of oscillation is \(0.05 \mathrm{~s}\) and the velocity of the wave is \(300 \mathrm{~m} / \mathrm{s}\). What is the phase difference between the oscillations of two points? (a) \(\pi\) (b) \(\frac{\pi}{6}\) (c) \(\frac{\pi}{3}\) (d) \(\frac{2 \pi}{3}\)

A particle is having kinetic energy \(1 / 3\) of the maximum value at a distance of \(4 \mathrm{~cm}\) from the mean position. Find the amplitude of motion. (a) \(2 \sqrt{6} \mathrm{~cm}\) (b) \(2 / \sqrt{6} \mathrm{~cm}\) (c) \(\sqrt{2} \mathrm{~cm}\) (d) \(6 / \sqrt{2}\)

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