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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?

Zain

7 Years agoGrade 12th pass

3 Answers

Eshan

7 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

7 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

5 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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