Question

a particle is in shm with amplitude A and has a maximum vel v (1)at what displacement will the velocity be v/2 (2)what is the velocity at displacement A/2

Answer 1

Dear Student,

The velocity of particle in SHM is given by,
$$\begin{gathered} v=\omega \sqrt{A^2-x^2} \\ \omega -angularfrequencyand \\ Aistheamplitude, \\ x-displacementfrommean. \\ Maximumvelocityisattainatx=0 \\ atx=0,V=\omega A. \\ \frac{V}{\omega }=A \\ 1)V_0=\frac{v}{2},x=? \\ \frac{v}{2}=\omega \sqrt{A^2-x^2} \\ \frac{v^2}{4{\omega }^2}=A^2-x^2 \\ {x}^2=A^2-\frac{v^2}{4{\omega }^2}=A^2-\frac{A^2}{4}=\frac{3A^2}{4} \\ Thereforex=\pm \frac{\sqrt{3}A}{2}, \\ 2)V=?atx=A/2. \\ {v}_o=\omega \sqrt{A^2-{\frac{A}{4}}^2}=\omega \pm \frac{\sqrt{3}}{2}A=\pm \omega \frac{\sqrt{3}}{2}\frac{v}{\omega }=\pm \frac{\sqrt{3}V}{2}(velocityatx=A/2) \end{gathered}$$
Regards.