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Chapter 10: Problem 1338

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

Short Answer

Expert verified
The distance traveled by a particle performing simple harmonic motion (SHM) during a time interval equal to its periodic time is \(2 \times A\). The correct answer is (B) \(2 \mathrm{~A}\).

Step by step solution

01

Recall the properties of SHM

Simple harmonic motion (SHM) is a type of oscillatory motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction. Some important properties of SHM are: 1. The particle oscillates between two extreme points symmetrically placed around the mean position. 2. The maximum displacement from the mean position is called the amplitude (A). 3. The time taken by the particle to complete one full oscillation is called the periodic time (T).
02

Understand the meaning of periodic time

Periodic time (T) is the time taken by the particle to complete one full oscillation or cycle. It is also the time interval after which the motion of the particle repeats itself in a periodic manner. The particle will be back to its initial position after completing one cycle.
03

Calculate the distance traveled during the periodic time

During one cycle of periodic time, the particle moves from one extreme position to the other and back to the initial extreme position. Since the amplitude represents the maximum displacement from the mean position, during one cycle, the particle will cover a distance equal to 2 times the amplitude (once while going from one extreme to the other and once while returning to its initial position). Therefore, the distance traveled by a particle performing SHM during a time interval equal to its periodic time is \(2 \times A\). The correct answer is (B) \(2 \mathrm{~A}\).

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Most popular questions from this chapter

The amplitude for a S.H.M. given by the equation $\mathrm{x}=3 \sin 3 \mathrm{pt}+4 \cos 3 \mathrm{pt}\( is \)\ldots \ldots \ldots \ldots \mathrm{m}$ (A) 5 (B) 7 (C) 4 (D) \(3 .\)

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If the mass of 1 mole of air is \(29 \times 10^{-3} \mathrm{~kg}\), then the speed of sound in it at STP is $(\gamma=7 / 5) .\left\\{\mathrm{T}=273 \mathrm{~K}, \mathrm{P}=1.01 \times 10^{5} \mathrm{~Pa}\right\\}$ (A) \(270 \mathrm{~m} / \mathrm{s}\) (B) \(290 \mathrm{~m} / \mathrm{s}\) (C) \(330 \mathrm{~m} / \mathrm{s}\) (D) \(350 \mathrm{~m} / \mathrm{s}\)
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