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Assertion The amplitude of a particle executing SHM with a frequency of \(60 \mathrm{~Hz}\) is \(0.01 \mathrm{~m}\). The maximum value of acceleration of the particle is \(\pm 144 \pi^{2} \mathrm{~ms}^{-2}\). Reason Acceleration amplitude \(=\omega^{2} A\),where \(A\) is displacement amplitude.

Short Answer

Expert verified
The assertion and the reason are correct and consistent.

Step by step solution

01

Understanding the Problem

We need to verify the assertion and the reason provided. The assertion states that the amplitude is 0.01 m, frequency is 60 Hz, and maximum acceleration is \( \pm 144 \pi^2 \ m/s^2 \). The reason gives a formula for acceleration amplitude as \( \omega^2 A \).
02

Convert Frequency to Angular Frequency

The frequency \( f \) is given as 60 Hz. We can calculate the angular frequency \( \omega \) using the formula \( \omega = 2\pi f \). Thus, \( \omega = 2\pi \times 60 \).
03

Calculate \(\omega\)

Substitute the known frequency into the angular frequency formula:\[ \omega = 2 \times \pi \times 60 = 120 \pi \text{ rad/s} \]
04

Find the Theoretical Maximum Acceleration

Using the reason's formula \( a_{max} = \omega^2 A \), substitute \( \omega = 120\pi \) and \( A = 0.01 \ m \):\[ a_{max} = (120\pi)^2 \times 0.01 = 14400\pi^2 \times 0.01 = 144\pi^2 \ ms^{-2} \].
05

Compare Theoretical and Given Values

The maximum acceleration calculated \( 144 \pi^2 \ m/s^2 \) matches the given value in the assertion \( \pm 144 \pi^2 \ ms^{-2} \). Thus, the assertion aligns with the reason provided.

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Key Concepts

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

Angular Frequency
In the realm of Simple Harmonic Motion (SHM), angular frequency plays a vital role. Angular frequency, often denoted as \( \omega \), offers insight into how quickly an object oscillates through its motion. It's an extension of the concept of frequency that we often encounter in regular wave phenomena.

To find the angular frequency from a regular frequency, we multiply by \( 2\pi \) which arises from the fact that one full cycle of oscillation corresponds to \( 2\pi \) radians. So, the formula to calculate angular frequency is \( \omega = 2\pi f \).

  • Angular frequency \( \omega \) tells us how rapidly the particle moves through its oscillation cycle in radians per second.
  • It's closely linked to the frequency of the oscillation, \( f \), measured in cycles per second or Hertz (Hz).
For instance, in our problem where the frequency \( f \) is given as 60 Hz, it converts into angular frequency by \( \omega = 2\pi \times 60 = 120\pi \text{ rad/s} \). This tells us the rate at which the particle progresses through its oscillation measured in radians over time.
Maximum Acceleration
Maximum acceleration in SHM is a key indicator of how swiftly the velocity of an oscillating object can change at its extreme points. It describes the peak magnitude of acceleration that the particle experiences as it moves through its path. In SHM, maximum acceleration occurs at the maximum displacement from the equilibrium position.

  • The maximum acceleration \( a_{max} \) is calculated using the formula: \( a_{max} = \omega^2 A \), where \( A \) is the displacement amplitude, and \( \omega \) is the angular frequency.
  • It's important to understand that \( a_{max} \) represents the highest point of acceleration, typically found at the turns (extremes) of the motion.
As detailed in our example, using \( \omega = 120\pi \) rad/s and \( A = 0.01 \) m, the maximum acceleration computes to \( 144\pi^2 \text{ ms}^{-2} \), aligning perfectly with the assertion that suggests \( \pm 144\pi^2 \text{ ms}^{-2} \). This reinforces the knowledge that the amplitude and angular frequency are essential in determining max acceleration.
Displacement Amplitude
Displacement amplitude is a fundamental concept in SHM, reflecting the maximum extent of oscillation from the central, or equilibrium, position. It represents the furthest point reached by the oscillating particle. In essence, it's a measure of how far from equilibrium the particle travels during its motion.

  • Denoted as \( A \), displacement amplitude directly influences the energy of the system, with larger amplitudes indicating more energy.
  • It's measured in linear units such as meters (m), showing the farthest reach from the center.
In our situation, the displacement amplitude is given as \( 0.01 \) meters. This value is crucial because it not only affects the maximum speed and acceleration but also determines the total energy contained in the SHM system. The connection between the maximum acceleration and displacement amplitude is established through the formula \( a_{max} = \omega^2 A \). Understanding this relation helps in predicting the dynamics of the particle’s motion.

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Most popular questions from this chapter

One end of a spring of force constant \(k\) is fixed to a vertical wall and the other to a block of mass \(m\) resting on a smooth horizontal surface. There is another wall at a distance \(x_{0}\) from the block. The spring is then compressed by \(2 x_{0}\) and then released. The time taken to strike the wall iv (a) \(\frac{1}{6} \pi \sqrt{\frac{k}{m}}\) (b) \(\sqrt{\frac{k}{m}}\) (c) \(\frac{2 \pi}{3} \sqrt{\frac{m}{k}}\) (d) \(\frac{\pi}{4} \sqrt{\frac{k}{m}}\)

A particle of mass \(m\) executes simple harmonic motion with amplitude \(a\) and frequency \(v\). The average kinetic energy during its motion from the position of equilibrium to the end is \(\quad\) (a) \(2 \pi^{2} m a^{2} v^{2}\) (b) \(\pi^{2} m a^{2} v^{2}\) (c) \(\frac{1}{4} \pi^{2} m a^{2} v^{2}\) (d) \(4 \pi^{2} m a^{2} v^{2}\)

A ball of mass \((\mathrm{m}) 0.5 \mathrm{~kg}\) is attached to the end of a string having length \((L) 0.5 \mathrm{~m}\). The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that string can bear is \(324 \mathrm{~N}\).The maximum possible value of angular velocity of ball (in \(\mathrm{rad} / \mathrm{s}\) ) is (a) 9 (b) 18 (c) 27 (d) 36

The displacement of a particle is represented by the equation \(y=3 \cos \left(\frac{\pi}{4}-2 \omega t\right)\). The motion of the particle is [NCERT Exemplar] (a) simple harmonic with period \(2 \mathrm{p} / \omega\) (b) simple harmonic with period \(\pi / \omega\) (c) periodic but not simple harmonic (d) non-periodic

A mass \(1 \mathrm{~kg}\) suspended from a spring whose force constant is \(400 \mathrm{Nm}^{-1}\), executes simple harmonic oscillation. When the total energy of the oscillator is \(2 \mathrm{~J}\), the maximum acceleration experienced by the mass will be (a) \(2 \mathrm{~ms}^{-2}\) (b) \(4 \mathrm{~ms}^{-2}\) (c) \(40 \mathrm{~ms}^{-2}\) (d) \(400 \mathrm{~ms}^{-2}\)

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