MCQ
Two simple harmonic motions of angular frequency $100$ and $1000\,\,rad\,s^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is
- A$1:10^3$
- B$1:10^4$
- C$1:10$
- ✓$1:10^2$
✓
Answer
Correct option: D.
$1:10^2$
d
$\omega_{1}=100 \mathrm{rad} \mathrm{s}^{-1} ; \omega_{2}=1000 \mathrm{rad} \mathrm{s}^{-1}$
$\omega_{1}=100 \mathrm{rad} \mathrm{s}^{-1} ; \omega_{2}=1000 \mathrm{rad} \mathrm{s}^{-1}$
Maximum acceleration of $(1)=-\omega_{1}^{2} A$
Maximum acceleration of $(2)=-\omega_{2}^{2} A$
$\therefore \frac{\operatorname{accln}(1)}{\operatorname{accln}(2)}=\frac{\omega_{1}^{2}}{\omega_{2}^{2}}=\frac{(100)^{2}}{(1000)^{2}}=\frac{1}{100}$
$a(1): a(2)=1: 100$.
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