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A particle executing SHM has a maximum speed of \(30 \mathrm{~cm} / \mathrm{s}\) and a maximum acceleration of \(60 \mathrm{~cm} / \mathrm{s}^{2}\). The period of oscillation is [NCERT Exemplar] (a) \(\pi \mathrm{s} \quad\) (b) \(\frac{\pi}{2} 5\) (c) \(2 \pi \mathrm{s} \quad\) (d) \(\frac{\pi}{t} s\)

Short Answer

Expert verified
The period of oscillation is \( \pi \text{ s} \). Answer: (a) \( \pi \mathit{s} \).

Step by step solution

01

Formula for Maximum Speed in SHM

The maximum speed \( v_{max} \) of a particle undergoing Simple Harmonic Motion (SHM) is given by \( v_{max} = A \omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency.
02

Formula for Maximum Acceleration in SHM

The maximum acceleration \( a_{max} \) in SHM is given by the formula \( a_{max} = A \omega^2 \).
03

Relating maximum speed and acceleration

We know that \( v_{max} = 30 \text{ cm/s} \) and \( a_{max} = 60 \text{ cm/s}^2 \). Substituting these into the formulas gives:1. \( A \omega = 30 \)2. \( A \omega^2 = 60 \).
04

Solving for Angular Frequency \( \omega \)

To find \( \omega \), divide the equation for \( a_{max} \) by the equation for \( v_{max} \):\[\frac{A \omega^2}{A \omega} = \frac{60}{30}\]Simplifying this gives:\[\omega = 2 \text{ s}^{-1}\]
05

Relation between Angular Frequency and Period

The angular frequency \( \omega \) is related to the period \( T \) by the formula: \( \omega = \frac{2\pi}{T} \).Substituting \( \omega = 2 \) into this equation gives:\[ 2 = \frac{2\pi}{T} \]
06

Solving for the Period \( T \)

Rearrange the equation \( 2 = \frac{2\pi}{T} \) to solve for \( T \):\[ T = \frac{2\pi}{2} = \pi \text{ s} \].

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Key Concepts

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

Maximum Speed in Simple Harmonic Motion
In Simple Harmonic Motion (SHM), a particle exhibits a periodic back-and-forth motion. Understanding the concept of maximum speed is essential to grasp the dynamics of SHM. The maximum speed occurs as the particle passes through the equilibrium position, where it momentarily does not experience any force from the restoring medium.

To calculate the maximum speed in SHM, we use the formula:\[ v_{max} = A \omega \]where:
  • \( v_{max} \) is the maximum speed.
  • \( A \) is the amplitude, which is the maximum displacement from the equilibrium position.
  • \( \omega \) is the angular frequency.
By knowing the amplitude and the angular frequency, you can easily determine how fast the particle moves at its maximum speed. An important insight here is that the larger the amplitude or the higher the angular frequency, the greater the maximum speed. This makes intuitive sense because a pendulum will swing faster if it swings through a greater distance in the same time period!
Maximum Acceleration in Simple Harmonic Motion
Acceleration in SHM is another crucial factor, especially its maximum value. This indicates how quickly a particle can change its velocity. The acceleration reaches its peak when the particle is at the endpoints of its motion, where the force pulling it back towards equilibrium is greatest.

The formula for calculating maximum acceleration is:\[ a_{max} = A \omega^2 \]where:
  • \( a_{max} \) is the maximum acceleration.
  • \( A \) is the amplitude.
  • \( \omega \) is the angular frequency.
This equation shows that the maximum acceleration depends on both the amplitude and the square of the angular frequency. Therefore, both a larger amplitude and a higher angular frequency will increase the maximum acceleration, making the particle subject to greater forces and moving more rapidly towards the equilibrium position when displaced.
Angular Frequency in Simple Harmonic Motion
Angular frequency is a fundamental concept in SHM, representing how quickly a particle oscillates back and forth through its equilibrium point. It is linked directly to the oscillation period and indirectly to other properties like speed and acceleration.

The angular frequency \( \omega \) is given by:\[ \omega = \frac{2\pi}{T} \]where:
  • \( \omega \) is the angular frequency.
  • \( T \) is the period of oscillation, which is the time taken to complete one full cycle of motion.
Angular frequency provides an easy way to understand a system's dynamics. A larger \( \omega \) signifies faster oscillations, while a smaller value indicates slower oscillations. In simple harmonic oscillatory systems such as springs or pendulums, \( \omega \) plays a key role in determining the characteristics of motion, including the speed and acceleration detailed in the previous sections. Understanding \( \omega \) helps predict and analyze the behavior of oscillating systems effectively.

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Most popular questions from this chapter

A particle is executing simple harmonic motion with an amplitude \(A\) and time period \(T\). The displacement of the particle after \(2 T\) period from its initial position is (a) \(\underline{A}\) (b) \(4 \mathrm{~A}\) (c) \(8 \mathrm{~A}\) (d) zero

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.

Two pendulums begin to swing simultaneously. The first pendulum makes 9 full oscillations when the other makes \(7 .\) The ratio of lengths of the two pendulums is (a) \(9 / 7\) (b) \(7 / 9\) (c) \(49 / 81\) (d) \(81 / 49\)

A particle is vibrating in a simple harmonic motion with and amplitude of \(4 \mathrm{~cm}\). At what displacement from the equilibrium position is its energy half potential and half kinetic? (a) \(1 \mathrm{~cm}\) \(\begin{array}{ll}\text { (b) } \sqrt{2} \mathrm{~cm} & \text { (c) } 3 \mathrm{~cm}\end{array}\) (d) \(2 \sqrt{2} \mathrm{~cm}\)

On a smooth inclined plane, a body of mass \(M\) is attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant \(k\), the period of oscillation of the body (assuming the springs as massless) is (a) \(2 \pi[M / 2 k]^{1 / 2}\) (b) \(2 \pi[2 M / k]^{1 / 2}\) (c) \(2 \pi[M g \sin \theta / 2 k]^{1 / 2}\) (d) \(2 \pi[2 M g / k]^{1 / 2}\)

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