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A block of mass m is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant k, and the amplitude of the SHM is A. The block has v = 0, and x = +A at t = 0. It first reaches x = 0 when t=T/4, where T is the period of the motion.a) In terms of T, what is the time t when the block first reaches x=A/2?
b) The block has its maximum speed when t=T/4. What is the value of t when the speed of the block first reaches the value vmax/2?

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Answer:

[tex]\dfrac{T}{6}[/tex]

[tex]\dfrac{T}{12}[/tex]

Explanation:

Equation of motion is given by

[tex]x=Acos\omega t[/tex]

When,

[tex]x=\dfrac{A}{2}[/tex]

[tex]\dfrac{A}{2}=Acos\omega t\\\Rightarrow cos\omega t=\dfrac{1}{2}\\\Rightarrow \omega t=cos^{-1}\dfrac{1}{2}\\\Rightarrow \omega t=\dfrac{\pi}{3}\\\Rightarrow \dfrac{2\pi}{T}t=\dfrac{\pi}{3}\\\Rightarrow t=\dfrac{T}{6}[/tex]

Time taken is [tex]\dfrac{T}{6}[/tex]

Velocity is given by

[tex]v=\dfrac{dx}{dt}=-A\omega sin\omega t[/tex]

Speed becomes half which means,

[tex]sin\omega t=\dfrac{1}{2}\\\Rightarrow \omega t=\dfrac{\pi}{6}\\\Rightarrow \dfrac{2\pi}{T}t=\dfrac{\pi}{6}\\\Rightarrow t=\dfrac{T}{12}[/tex]

Time taken is [tex]\dfrac{T}{12}[/tex]

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