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A particle P describes S.H.M. of amplitude \(1 \mathrm{~m}\). In performing one complete oscillation, P travels a distance: (a) \(2 \mathrm{~m}\) (b) 0 (c) \(4 \mathrm{~m}\) (d) \(-2 \mathrm{~m}\).

Short Answer

Expert verified
Option (c), 4 m.

Step by step solution

01

Understand the Problem

A particle P undergoes simple harmonic motion (S.H.M.) with an amplitude of 1 meter. The goal is to find the total distance P travels in one complete oscillation.
02

Calculate the Distance for Half an Oscillation

In one half of an oscillation, the particle moves from one extreme point to the other (from +1 m to -1 m). This distance is given by twice the amplitude: \[ \text{Distance} = 2 \times 1 \text{ m} = 2 \text{ m} \]
03

Calculate the Full Oscillation

In a complete oscillation, the particle moves from one extreme to the other and then back again to its starting point. Therefore, the distance for a full oscillation is twice the distance of a half oscillation: \[ \text{Total Distance} = 2 \times 2 \text{ m} = 4 \text{ m} \]
04

Select the Correct Option

From the given options, the correct distance traveled by the particle in one complete oscillation is 4 meters, which corresponds to option (c).

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Key Concepts

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

Amplitude
Amplitude is a key concept in simple harmonic motion (S.H.M.). It represents the maximum displacement of the particle from its mean or equilibrium position. In other words, it's the farthest distance the particle moves away from the center during its motion.
For example, if a particle has an amplitude of 1 meter, it means the particle oscillates 1 meter to the left and 1 meter to the right of its equilibrium point. The total range, therefore, would be 2 meters.
To remember:
  • Amplitude is always a positive value.
  • It measures the peak of the oscillation, not the total distance traveled.
  • In the exercise given, the amplitude is 1 meter.
Oscillation
Oscillation refers to the repeated back and forth motion of a particle about its equilibrium position. In simple harmonic motion, this movement is periodic and predictable.
An oscillation has two key parts:
  • Half Oscillation: The particle moves from one extreme (e.g., +1 meter) to the other extreme (e.g., -1 meter).
  • Full Oscillation: The particle returns to its starting point after completing a half oscillation.
In the exercise, to calculate the distance for one full oscillation:
1. The distance for a half oscillation from +1 meter to -1 meter is twice the amplitude:\[\text{Distance} = 2 \times 1 \text{ m} = 2 \text{ m}\]
2. The distance for a full oscillation is twice the distance of a half oscillation: \[\text{Total Distance} = 2 \times 2 \text{ m} = 4 \text{ m}\]
Hence, the particle moves 4 meters in a complete oscillation.
Distance Calculation
Distance calculation is crucial in understanding the total distance a particle travels in simple harmonic motion. Let's break it down further.
Given the amplitude (A) is 1 meter:
1. The particle moves from the mean position to the maximum positive amplitude (+1 meter).
2. It then moves from +1 meter to the maximum negative amplitude (-1 meter).
3. This back and forth motion from +1 to -1 meter constitutes a half oscillation. The distance for this is twice the amplitude: \[\text{Distance}_{\text{half}} = 2 \times A = 2 \times 1 \text{ m} = 2 \text{ m}\]
4. To complete one full oscillation, the particle returns to its starting point, covering another half oscillation. Thus, the total distance for a full oscillation is: \[\text{Distance}_{\text{full}} = 2 \times \text{Distance}_{\text{half}} = 2 \times 2 \text{ m} = 4 \text{ m}\]
So, the particle travels 4 meters in one complete oscillation.

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Most popular questions from this chapter

A particle of mass \(m\) is suspended from a ceiling by a light elastic string, of natural length \(a\) and modulus \(12 m g .\) When the particle hangs at rest find the extension in the string. The particle is then pulled down vertically a distance \(x\) and released. If the particle just reaches the ceiling, find: (a) the value of \(x\), (b) the maximum speed and the maximum acceleration during the motion.

A particle performing S.H.M. has a speed of \(4 \mathrm{~ms}^{-1}\) when it is \(1 \mathrm{~m}\) from the centre. If the amplitude is \(3 \mathrm{~m}\) what is the period of oscillation? (a) \(\sqrt{2} \pi\) (b) \(\frac{\pi}{\sqrt{2}}\) (c) \(\frac{\pi}{2}\) (d) \(\frac{\pi}{2 \sqrt{2}}\).

A particle is moving with linear simple harmonic motion. Its speed is maximum at a point \(C\) and is zero at a point A. P and \(Q\) are two points on CA such that \(4 \mathrm{CP}=\mathrm{CA}\) while the speed at \(\mathrm{P}\) is twice the speed at \(\mathrm{Q}\). Find the ratio of the accelerations at \(\mathrm{P}\) and \(Q\). If the period of one oscillation is 10 seconds find, correct to the first decimal place, the least time taken to travel between \(\mathrm{P}\) and \(\mathrm{Q}\).

Prove that, if a particle moving with linear simple harmonic motion of amplitude \(a\) has velocity \(v\) when distant \(x\) from the centre of its path, then \(v=\omega \sqrt{a^{2}-x^{2}}\) where \(\omega\) is a constant. A point travelling with linear S.H.M. has speeds \(3 \mathrm{~ms}^{-1}\) and \(2 \mathrm{~ms}^{-1}\) when distant \(1 \mathrm{~m}\) and \(2 \mathrm{~m}\) respectively from the centre of oscillations. Calculate the amplitude, the periodic time and the maximum velocity.

A particle of mass 10 grammes is moving along a straight line with simple harmonic motion. The particle has speeds of 9 centimetres per second and 6 centimetres per second at \(P\) and \(Q\) respectively, whose distances from the centre of oscillation are 1 centimetre and 2 centimetres respectively. Calculate the greatest speed and the greatest acceleration of the particle. If the points \(\mathrm{P}\) and \(\mathrm{Q}\) are on the same side of the centre of oscillation, calculate: (i) the shortest time taken by the particle to move from P to Q. (ii) the work done during this displacement.

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