# 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?

Eshan
3 years ago
Dear student,

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

$x=Acos\omega t$
Hence for$x=A/2$,
$A/2=Acos\omega t\implies \omega t=\dfrac{\pi}{3}$
$\implies t=\dfrac{\pi}{3\omega}$
Since$T=\dfrac{2\pi}{\omega}$,
$t=T/6$
Dev Kumar
37 Points
3 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
one year ago
Dear student,

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