Q. For a particle in SHM, if the amplitude of displacement is $a$ and the amplitude of velocity is $v$ , then what will be the amplitude of acceleration?
NTA AbhyasNTA Abhyas 2022
Solution:
The equation for position of a particle under SHM, $x=asin\left(\omega t\right)$ ,
$\Rightarrow v=a\omega cos\left(\omega t\right)$ ,
maximum velocity, $v=\omega a$
$\text{Acceleration}=-a\left(\omega \right)^{2}sin\left(\omega t\right)$ ,
maximum acceleration, $f=\left(\omega \right)^{2}a\Rightarrow f=\left(\frac{v}{a}\right)^{2}a\Rightarrow f=\frac{v^{2}}{a}$
$\Rightarrow v=a\omega cos\left(\omega t\right)$ ,
maximum velocity, $v=\omega a$
$\text{Acceleration}=-a\left(\omega \right)^{2}sin\left(\omega t\right)$ ,
maximum acceleration, $f=\left(\omega \right)^{2}a\Rightarrow f=\left(\frac{v}{a}\right)^{2}a\Rightarrow f=\frac{v^{2}}{a}$