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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?


2 years ago

Eshan
2095 Points
							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$

2 years ago
Dev Kumar
37 Points
							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---->ADraw 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
605 Points
							Dear student,Please find the attached solution to your problem. Assume x = 0 as t = 0 and T be the period of SHMTime taken to travel between extreme position, x = A and initial position, x = 0 is T/4Now, x = A sin(2π/T * t)A/2 = A sin(2π/T * t)or, t = T/12Hence, time taken to reach x = A/2 from x = A isT/4 – T/12 = T/6 Hope it helpsThanks and regards,Kushagra

4 months ago
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