Q. A body executing simple harmonic motion has a maximum acceleration equal to 24 $ms^{- 2}$ and maximum velocity of 16 $ms^{- 1}$, the amplitude of the simple harmonic motion is
A) $\frac{1024}{9}$ m
B) $\frac{32}{3}$ m
C) $\frac{64}{9}$ m
D) $\frac{3}{32}$m
Explanation
(b) Maximum acceleration is given by
$= a\omega^{2} = 24\ ms^{- 2}$ …(i)
Maximum velocity= $a\omega = 16\ ms^{- 2}$ …(ii)
Dividing Eq. (i) by Eq. (ii)
$\frac{a\omega^{2}}{a\omega} = \omega = \frac{24}{16} = \frac{6}{4} = \frac{3}{2}$
Now putting the value of $\omega$ in Eq. (ii), we get
$a \times \frac{3}{2} = 16$
$a = \frac{32}{3}$ m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.