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A simple harmonic oscillator has an amplitude A and time period T the time required by it to travel from x=A to x=A/2 is?

A simple harmonic oscillator has an amplitude A and time period T the time required by it to travel from x=A to x=A/2 is?

Grade:12th pass

3 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear student,

Since the particle starts from x=A, the equation of motion can be written as

x=Acos\omega t
Hence forx=A/2,
A/2=Acos\omega t\implies \omega t=\dfrac{\pi}{3}
\implies t=\dfrac{\pi}{3\omega}
SinceT=\dfrac{2\pi}{\omega},
t=T/6
Dev Kumar
37 Points
4 years ago
You could've been answered the question with the help of concept of phaser diagram.
In this, draw a circle of radius A on a Cartesian sheet, and on y-axis mark a point A/2 on +y-axis,and at the end point of circle mark point A 
Particle shm => A----->A/2 ; can be considered as A/2---->A
Draw a line  from A/2 projecting on the circle, join the origin and projection, take angle along y-axis and mark it as μ.
Now cos μ= A/2A= 1/2 => μ=π/3 
We know that,
                           μ/t= 2π/T
=> π/3t=2π/T
=> t= T/6
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your problem.
 
Assume x = 0 as t = 0 and T be the period of SHM
Time taken to travel between extreme position, x = A and initial position, x = 0 is T/4
Now, x = A sin(2π/T * t)
A/2 = A sin(2π/T * t)
or, t = T/12
Hence, time taken to reach x = A/2 from x = A is
T/4 – T/12 = T/6
 
Hope it helps
Thanks and regards,
Kushagra

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