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

`(T)/(6)`

`(T)/(12)`

Step-by-step explanation:

Equation of motion is given by

`x=Acos\omega t`

When,

`x=(A)/(2)`

`(A)/(2)=Acos\omega t\\\Rightarrow cos\omega t=(1)/(2)\\\Rightarrow \omega t=cos^(-1)(1)/(2)\\\Rightarrow \omega t=(\pi)/(3)\\\Rightarrow (2\pi)/(T)t=(\pi)/(3)\\\Rightarrow t=(T)/(6)`

Time taken is
`(T)/(6)`

Velocity is given by

`v=(dx)/(dt)=-A\omega sin\omega t`

Speed becomes half which means,

`sin\omega t=(1)/(2)\\\Rightarrow \omega t=(\pi)/(6)\\\Rightarrow (2\pi)/(T)t=(\pi)/(6)\\\Rightarrow t=(T)/(12)`

Time taken is
`(T)/(12)`