Question

The amplitude of a particle executing SHM is 2 cm and its frequency is 50 cycles/s.Its maximum acceleration will be

• A. $1972m{s}^{-2}$
• B. $19.72m{s}^{-2}$
• C. $1.972m{s}^{-2}$
• D. zero

Answer 1

The correct option is A $1972m{s}^{-2}$

We know that $a=-{\omega }^{2}x$ and maximum x is amplitude A. Thus, maximum magnitude of acceleration, $a={\omega }^{2}A$
Now, $\omega =2\pi f=100\pi$ (f=50)
Thus, maximum $a=10000{\pi }^2\times 0.02$ (A=2 cm = 0.02 m)
Thus, $a=1972m/s^2$ nearly