Question

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?

Answer 1

Answer:

$\dfrac{T}{6}$

$\dfrac{T}{12}$

Explanation:

Equation of motion is given by

$x=Acos\omega t$

When,

$x=\dfrac{A}{2}$

$\dfrac{A}{2}=Acos\omega t\\\Rightarrow cos\omega t=\dfrac{1}{2}\\\Rightarrow \omega t=cos^{-1}\dfrac{1}{2}\\\Rightarrow \omega t=\dfrac{\pi}{3}\\\Rightarrow \dfrac{2\pi}{T}t=\dfrac{\pi}{3}\\\Rightarrow t=\dfrac{T}{6}$

Time taken is $\dfrac{T}{6}$

Velocity is given by

$v=\dfrac{dx}{dt}=-A\omega sin\omega t$

Speed becomes half which means,

$sin\omega t=\dfrac{1}{2}\\\Rightarrow \omega t=\dfrac{\pi}{6}\\\Rightarrow \dfrac{2\pi}{T}t=\dfrac{\pi}{6}\\\Rightarrow t=\dfrac{T}{12}$

Time taken is $\dfrac{T}{12}$